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TITRIMETRIC ANALYSIS

1. Define the following terms

Volumetric analysis

Volumetric  analysis refers to quantitative chemical analysis carried out by determining the volume of a solution of accurately known concentration which is required to react quantitatively with a measured volume of a solution of the substance to be determined.

Aliquot

Titrand:the substance being titrated

Titrant:the reagent of known concentration

Titre:The volume  used of the substance titrated

Titration:The process of adding the standard solution until the reaction is just complete

Indicator: Indicator is a chemical substance which changes colour at the end point.

End Point:The stage during titration at which the reaction is just complete is known as the end point of titration

Equivalent point;The point at which this occurs is called the equivalence point or the theoretical (or stoichiometric) end point

Explain  the differences between

1. 1. End point and equivalence point?

End point is the point which indicates the completion of reaction by changing its color. It is indicated by the indicator. While equivalence point is the theoretical point at which the equivalent amount of titrand and titrant are added together.  It is a theoretical point.

1. Equivalence & endpoint in titrations?

The point at which the reaction b/w titrant & titrate is just complete is called equivalence point or theoretical point.

The point at which a clear visual change is observed after the reaction b/w titranta and titrate is practically complete is end point.

Thus there exists a very small difference b/w the end point & equivalence point.

3. Explain  the term  

Equivalent weight of an Acid?

Equivalent weight of an acid is defined as the number of parts by mass of an acid that is neutralised completely by one equivalent weight of base. Equivalent weight of an acid = molecular weight / basicity

b.Equivalent weight of a Base?

Equivalent weight of a base is defined as the number of parts by mass of a base that is required to neutralise completely one equivalent weight of an acid. Equivalent weight of a base = molecular weight / acidity

          EX: Equivalent weight of NaOH =40/1=40

1. Which indicator is used in the titration of sodium carbonate against hydrochloric acid and what is the colour change at the end point?

Methyl orange. The colour change is yellow to pinkish red

2. List any two advantages of  volumetric over gravimetric analysis

Volumetric analysis require  simpler apparatus, and are, generally, quickly performed; tedious and difficult separations can often be avoided.

3. State conditions necessary for analysis of a sample  by volumetric analysis

• . There must be a simple reaction which can be expressed by a chemical equation; the substance to be determined should react completely with the reagent in stoichiometric or equivalent proportions.
• The reaction should be relatively fast. (Most ionic reactions satisfy this condition.) In some cases the addition of a catalyst may be necessary to increase the speed of a reaction.
• There must be an alteration in some physical or chemical property of the solution at the equivalence point.
• An indicator should be available which, by a change in physical properties (colour or formation of a precipitate), should sharply define the end point of the reaction. If no visible indicator is available, the detection of the equivalence point can often be achieved by following the course of the titration by measuring

4. Name four types of volumetric analysis

The volumetric analysis can be classified into three types:

1. Simple titration
2. Back titration   
3. Double titrations

SIMPLE TITRATION The main aim of simple titration is to find the concentration of an unknown solution with the aid of the known concentration of another solution. Simple titration can again be classified into four different types:•

1. Acid-base titrations•
2. Redox titrations•
3. Precipitation titrations and•
4. Complexometric titrations
5. State two possible sources of errors in volumetric analysis

It can be either of end point error, misreading volumes, concentrations, faulty use of equipment, contaminated glass ware, etc.

6. State four advantages of titrimetric  as a method of analysis

• Volumetric analysis require  simpler apparatus,
• Is  generally quickly performed;
• Tedious and difficult separations can often be avoided.
• Require less skills

 

7. A solution of oxalic acid  was prepared by disolving 0.5g of oxalic acid (H₂C₂O2H₂O) in a 250cm3 volumetric flask.25cm3 of the prepared solution  required 30cm3 of the acidified potassium permanganete to complete the filtration . calculate

• Number of moles of exalic acid  solution in 240cm3 solution (H=1.C=12,O=16)
• Moles of oxalic acid in 25ml
• Moles of permanganete ions
• Concentration of permanganate ion solution in moles per liter

 

8. Define a standard  solution

 

A standard solution is a solution containing a precisely known concentration of an element or a substance. A known mass of solute is dissolved to make a specific volume. It is prepared using a standard substance, such as a primary standard. Standard solutions are used to determine the concentrations of other substances, such as solutions in titration.

The concentrations of standard solutions are normally expressed in units of moles per litre (mol/L, often abbreviated to M for molarity), moles per cubic decimetre (mol/dm3), kilomoles per cubic metre (kmol/m3) or in terms related to those used in particular titrations (such as titres A simple standard is obtained by the dilution of a single element or a substance in a soluble solvent with which it reacts. A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has high molecular weight. Some primary standards of titration of acids include sodium carbonate.

9. Distinguish between a primary and a secondary standard solution

A primary standard is a highly purified compound that serves as a reference material in titrations and in other analytical methods.

The accuracy of a method critically depends on the properties of the primary standard.

• List the  requirements for primary standard solution :

1. It should be available in pure and dry state.
2. Its composition should not be changed on storage. i.e., It should not be toxic, volatile, hygroscopic and deliquescent.
3. It should be highly soluble in water without any change in its composition.
4. It should have higher molecular and equivalent weight
5. Modest cost or relatively affordable .
6. Should have reasonable solubility in the titration medium.
7. Reasonably large molar mass so that the relative error associated with weighing the standard is minimized.

Very few compounds meet or even approach these criteria, and only a limited number of primary-standard substances are available commercially. As a consequence, less pure compounds must sometimes be used in place of a primary standard. The purity of such a secondary standard must be established by careful analysis. 

A secondary standard is a compound whose purity has been determined by chemical analysis. The secondary standard serves as the working standard material for titrations and for many other analyses.

 

10. Give  examples of primary standard solution :

Sodium carbonate (Na2CO3), Potassium dichromate (K2Cr2O7), oxalic acid (COOH)2.2H2O,  Mohr’s salt etc.

11. List two methods of preparing standard solution
12. Explain the importance of primary  and standard solutions

 A primary standard is an ultrapure compound that serves as the reference material for a titration or for another type of quantitative analysis.

Standard Solutions play a central role in  the determination of  the concentration of an analyte species.  Just like a primary  standard ,the standard solution is a reference guide to discover the molarity of unknown species.

Titration methods can be used to acquire the concentration of a standard solution. These involve using equipment such as a burette.

By comparing the absorbance of the sample solution at a specific wavelength to a series of standard solutions at differing known as concentrations of the analyse species, the concentration of the sample solution can be found via Beer’s Law. Any form of spectroscopy can be used in this way so long as the analyte species has substantial absorbance in the spectra.

Therefore, we must consider the desirable properties for such solutions, how they are prepared, and how their concentrations are expressed.

The ideal standard solution for a titrimetric method will:

1. Re sufficiently stable so that it is necessary to determine its concentration only once;
2. React rapidly with the analyte so that the time required between additions of reagent is minimized;
3. React more or less completely with the analyte so that satisfactory end points are realized;
4. Undergo a selective reaction with the analyte that can be described by a balanced equation.

Few reagents completely meet these ideals. The accuracy of a titration can be no better than the accuracy of the concentration of the standard solution used.

13. Explain the methods used  to establish  the concentration of   standard solutions

Two basic methods are used to establish the concentration of such solutions. The first is the direct method in which a carefully determined mass of a primary standard is dissolved in a suitable solvent and diluted to a known volume in a volumetric flask.

The second is by standardization in which the titrant to be standardized is used to titrate

(1) a known mass of a primary standard,

(2) a known mass of a secondary standard, or

(3) a measured volume of another standard solution.

A titrant that is standardized is sometimes referred to as a secondary standard solution. The concentration of a secondary-standard solution is subject to a larger uncertainty than is the concentration of a primary-standard solution. If there is a choice, then, solutions are best prepared by the direct method.

Many reagents, however, lack the properties required for a primary standard and, therefore, require standardization

13.Outline the various types  of  titration

There are many types of titration when considering goals and procedures. However, the most common types of titration in quantitative chemical analysis are redox titration and acid-base titration.

Titrations can be classified as:

1. Acid-base Titrations
2. Redox Titrations.
3. Precipitation Titrations.
4. Complexometric Titrations
5. Acid-Base Titration

The strength of an acid can be determined using a standard solution of a base. This process is called acidimetry. In the same way, the strength of a base can be found with the help of a standard solution of an acid, which is known as alkalimetry. Both titrations involve in the neutralization reaction of an alkali.

 Acid- base  titration is a quantitative analysis method to determine an acid’s or bases’ concentration by precisely neutralizing them with a standard solution of either acid or base of known concentration. It is monitored with the help of a pH indicator to know the development of the acid-base reaction.

HA+BOH→BA+H₂O

Acid + Alkali→Salt + Water

Or H⁺ + A^– + B⁺ + OH^– → B⁺ + A^– + H₂O

Or H⁺ + OH^– → H₂O

The acid-base titration is based on the reaction that neutralization is between a base or an acidic and analyte. In this type, a reagent is mixed with the sample solution until it reaches the required pH level. This type of titration majorly depends on the track change in pH or a pH meter.

1. Redox Titrations

The redox titration is also known as an oxidation-reduction reaction. In this type of titration, the chemical reaction takes place with a transfer of electrons in the reacting ions of aqueous solutions. The titrations are named after the reagent that is used in are as follows;

• Permanganate Titrations
• Dichromate Titrations
• Iodimetric and Iodometric Titrations

Permanganate Titrations

In this titration, the potassium permanganate is used as an oxidizing agent. It is maintained with the use of dilute sulphuric acid. Here is the equation.

 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]

               Or 

MnO₄^– + 8H⁺ + 5e^– → Mn²⁺+ 4H₂O

Further, the solution remains colourless before the endpoint. The potassium permanganate is used to estimate oxalic acid, ferrous salts, hydrogen peroxide, oxalates and more. While the solution of potassium permanganate is always standardized before it is used.

Dichromate Titrations

These are titrations in which, potassium dichromate is used as an oxidising agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. The potential equation is:

K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]

                           Or 

Cr₂O₇^2– + 14H⁺ + 6e^– → 2 Cr³⁺ + 7H₂O 

The solution of potassium dichromate can be directly used for titrations. It is mainly used for the estimation of ferrous salts and iodides.

Iodimetric and Iodometric Titrations

The reduction of free iodine to iodide ions and

oxidation of iodide ions to free occurs in these titrations.

l₂ + 2e^– → 2l^–……………. (reduction)

2l^– → I₂  + 2e^– ……………. (oxidation)

The solution is used as an indicator. Free iodine is used in the iodometric titration, while in the iodometric titration an oxidation agent is used to react to liberate free iodine.

1. Precipitation Titrations

The titration is based on the insoluble precipitate formation when the two reacting substances are brought into contact are called precipitation titration. For instance, when the solution of silver nitrate is used to a solution of ammonium thiocyanate or sodium chloride, it reacts and forms a white precipitate of silver thiocyanate or silver chloride.

 

AgNO₃ + NaCl → AgCl + NaNO₃

AgNO₃ + NH₄CNS → AgCNS + NH₄NO₃

 

1. Complexometric Titrations

The complexometric titration is where an undissociated complex is formed at an equivalence point. It is greater than the precipitation titrations, and there will be no error due to co-precipitations.

Hg²⁺ + 2SCN^– → Hg(SCN)₂

Ag⁺ + 2CN^– → [Ag(CN)₂]^–

Ethylenediaminetetraacetic acid (EDTA) is an important reagent that forms complexes with metals.

1. Explain  Acid-base reaction

An acid-base reaction is one in which a hydrogen ion, H⁺, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion.

The reaction between an acid and a base is called an acid-base reaction or a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base cancel each other’s chemistry to produce a rather innocuous substance—water. In fact, the general acid-base reaction is

acid + base ⟶ water + salt

where the term salt is used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. In chemistry, the word salt refers to more than just table salt. For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is

HCl(aq) + KOH(aq) ⟶ H₂O(ℓ) + KCl(aq)

where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)₂(aq), additional molecules of HCl and H₂O are required to balance the chemical equation:

2 HCl(aq) + Mg(OH)₂(aq) ⟶ 2 H₂O(ℓ) + MgCl₂(aq)

Here, the salt is MgCl₂. This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid. There are acid-base reactions that do not follow the “general acid-base” equation given above.  For example, , the balanced chemical equation for the reaction between HCl(aq) and NH₃(aq) is

HCl(aq) + NH₃(aq) ⟶ NH₄Cl(aq)

Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)₃(s) still proceeds according to the equation

3 HCl(aq) + Fe(OH)₃(s) ⟶ 3 H₂O(ℓ) + FeCl₃(aq)

Even though Fe(OH)₃ is not soluble. When one realizes that Fe(OH)₃(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids—the neutralization reaction produces products that are soluble and wash away. Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!

Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),

         HCl(aq) + NaOH(aq) ⟶ H₂O(ℓ) + NaCl(aq)

the complete ionic reaction is

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ H₂O(ℓ) + Na⁺(aq) + Cl⁻(aq)

The Na⁺(aq) and Cl⁻(aq) ions are spectator ions, so we can remove them to have

H⁺(aq) + OH⁻(aq) ⟶ H₂O(ℓ)

as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H₃O⁺(aq), we would write it as

H₃O⁺(aq) + OH⁻(aq) ⟶ 2H₂O(ℓ)

With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.

However, for the reaction between HCl(aq) and Cr(OH)₂(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:

2 H⁺(aq) + 2 Cl⁻(aq) + Cr(OH)₂(s) ⟶ 2 H₂O(ℓ) + Cr²⁺(aq) + 2 Cl⁻(aq)

The chloride ions are the only spectator ions here, so the net ionic equation is

2 H⁺(aq) + Cr(OH)₂(s) ⟶ 2 H₂O(ℓ) + Cr²⁺(aq)

2. State four conditions which necessitates  back titration in volumetric analysis

 

3. Outline the basic steps in back titration

A back titration, also called an indirect titration, is a titration method where the concentration of an analyte is determined by reacting it with a known amount of excess reagent.

Back titration allows the user to find the concentration of a reactant of unknown concentration by reacting it with an excess volume of another reactant of known concentration.

The resulting mixture is then titrated back with another, second reagent. The second titration’s result shows how much of the excess reagent was used in the first titration, thus allowing the original analyte’s concentration to be calculated.

 Back titrations can be used for many reasons, including: when the sample is not soluble in water, when the sample contains impurities that interfere with forward titration, or when the end-point is more easily identified than in forward titration.

A back titration is used when the molar concentration of an excess reactant is known, but the need exists to determine the strength or concentration of an analyte.

Back titration is typically applied in acid-base titrations:

1. When the acid or (more commonly) base is an insoluble salt (e.g., calcium carbonate)
2. Back titrations are applied, more generally, when the endpoint is easier to see than with a normal titration, (e.g., weak acid and weak base titration) or reactions which applies to some precipitation reactions.

 

13. Write the neutralization reactions between each acid and base.
14. a) HNO₃(aq) and Ba(OH)₂(aq)            

b)H₃PO₄(aq) and Ca(OH)_2(aq)

First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.

1. a) The expected products are water and barium nitrate, so the initial chemical reaction is

HNO₃(aq) + Ba(OH)₂(aq) ⟶ H₂O(ℓ) + Ba(NO₃)₂(aq)

To balance the equation, we need to realize that there will be two H₂O molecules, so two HNO₃ molecules are required:

2HNO₃(aq) + Ba(OH)₂(aq) ⟶ 2H₂O(ℓ) + Ba(NO₃)₂(aq)

This chemical equation is now balanced.

1. b) The expected products are water and calcium phosphate, so the initial chemical equation is

H₃PO₄(aq) + Ca(OH)₂(aq) ⟶ H₂O(ℓ) + Ca₃(PO₄)₂(s)

According to the solubility rules, Ca₃(PO₄)₂ is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:

2 H₃PO₄(aq) + 3 Ca(OH)₂(aq) ⟶ 6 H₂O(ℓ) + Ca₃(PO₄)₂(s)

This chemical equation is now balanced.

Write balanced chemical equations for the acid-base reactions described here:

1. a) the weak acidhydrogen hypochlorite reacts with water
2. b) a solution of barium hydroxide is neutralized with a solution of nitric acid

 The two reactants are provided, HOCl and H₂O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H⁺ from HOCl to H₂O to generate hydronium ions, H₃O⁺ and hypochlorite ions, OCl⁻.

HOCl(aq)+H2O(l)⇌OCl−(aq)+H3O+(aq)

A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.

The two reactants are provided, Ba(OH)₂ and HNO₃. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba²⁺) and the anion generated when the acid transfers its hydrogen ion (NO³⁻).

Ba(OH)2(aq)+2HNO3(aq)⟶Ba(NO3)2(aq)+2H2O(l)

Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. Hint: Consider the ions produced when a strong acid is dissolved in water.

H3O+(aq)+OH−(aq)⟶2H2O(l)

14.Oxalic acid, H₂C₂O₄(s), and Ca(OH)₂(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? The anion in oxalic acid is the oxalate ion, C₂O₄²⁻.

The products of the neutralization reaction will be water and calcium oxalate:

             H₂C₂O₄(s) + Ca(OH)₂(s) ⟶ 2 H₂O(ℓ) + CaC₂O₄(s)

Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.

What is the net ionic equation between HNO₃(aq) and Ti(OH)₄(s)?

            4 H⁺(aq) + Ti(OH)₄(s) ⟶ 4 H₂O(ℓ) + Ti⁴⁺(aq)

14.  A solution of sodium hydroxide contained 0.250 mol dm^-3. Using phenolphthalein indicator, titration of 25.0 cm³of this solution required 22.5 cm³of a hydrochloric acid solution for complete neutralisation.

(a) write the equation for the titration reaction. NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)

(b) what apparatus would you use to measure out (i) the sodium hydroxide solution? (ii) the hydrochloric acid solution?

  (i) pipette (ii) burette

(c) what would you rinse your apparatus out with before doing the titration ?

everything with distilled water, then pipette with a little of the NaOH(aq) and the burette with a little of the HCl(aq)

(d) what is the indicator colour change at the end-point? 

pink to colourless, the first drop of excess acid removes the pink alkaline colour of phenolphthalein

(e) calculate the moles of sodium hydroxide neutralised.

moles sodium hydroxide neutralised: 0.250 x 25.0/1000 

                    = 0.00625 mol NaOH

(remember: moles = molarity x volume in dm³ and its two rearrangements and 1 dm³ = 1000 cm³)

(f) calculate the moles of hydrochloric acid neutralised.

 moles HCl = moles NaOH (equation) = 0.00625 mol HCl (in 22.5 cm³)

(g) calculate the concentration of the hydrochloric acid in mol/dm³ 

(molarity).

 concentration hydrochloric acid = 0.00625 x 1000 ÷ 22.5 

= 0.278 mol dm^-3 (3sf)

(scaling up to 1 dm³ = 1000 cm³ to get the molarity)

Another way to work it out is 22.5 cm³ = 22.5 ÷ 1000 = 0.0225 dm³

Therefore molarity = 0.00625 ÷ 0.0225 =  0.278 mol dm^-3

15. A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?

Step 1: Determine [OH-]

Every mole of NaOH will have one mole of OH-. Therefore [OH-] = 0.5 M.

Step 2: Determine the number of moles of OH-

Molarity = number of moles/volume

Number of moles = Molarity x Volume

Number of moles OH- = (0.5 M)(0.025 L)

Number of moles OH- = 0.0125 mol

Step 3: Determine the number of moles of H+

When the base neutralizes the acid, the number of moles of H+ = the number of moles of OH-. Therefore, the number of moles of H+ = 0.0125 moles.

Step 4: Determine the concentration of HCl

  Every mole of HCl will produce one mole of H+; therefore, the number of moles of HCl = number of moles of H+.

Molarity = number of moles/volume

Molarity of HCl = (0.0125 mol)/(0.05 L)

Molarity of HCl = 0.25 M

Answer

The concentration of the HCl is 0.25 M.

Another Solution Method

The above steps can be reduced to one equation:

MacidVacid = MbaseVbase

Where

Macid = concentration of the acid

Vacid = volume of the acid

Mbase = concentration of the base

Vbase = volume of the base

This equation works for acid/base reactions where the mole ratio between acid and base is 1:1. If the ratio were different, as in Ca(OH)2 and HCl, the ratio would be 1 mole acid to 2 moles base. The equation would now be:

                       Macid Vacid = 2Mbase Vbase

For the example problem, the ratio is 1:1:

Macid Vacid = Mbase Vbase

Macid(50 ml)= (0.5 M)(25 ml)

Macid = 12.5 MmL

                50 ml

Macid = 0.25 M

 

16. In an experiment an unknown mass of anhydrous sodium carbonate was dissolved in water and the solution made up to 250cm³. 25cm³ of this solution neutralized 20cm³ of 0.25M nitric acid. (Na = 23.0 C = Na₂CO_3(aq)+ 2HNO_3(aq)→ 2NaNO_3(aq) + CO_2(g) + H₂O_(l)
    Calculate:
    1. Moles of Nitric acid used

Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + CO2(g) + H2O(l)

Mole ration 1 : 2

Moles of HNO₃ in 20cm³ = ²⁰/₁₀₀₀ x 0.25
= 0.005 moles

1. Moles of sodium carbonate in 25cm of the solution

Moles of Na₂CO₃ in 25cm³ = ½ of 0.005 moles
= 0.0025

1. Mass of unknown sodium carbonate used

If 25cm³ = 0.0025 moles
in 250cm³ = ?
250 x 0.0025
      25
= 0.025 moles
RFM of Na₂CO₃ = 106
1 mole of Na₂CO₃ = 106g
0.025 moles = ?
0.025 x 106
        1
= 2.65g of Na₂CO₃

17. A solution made from pure barium hydroxide contained 2.74 g in exactly 100 cm³of water. Using phenolphthalein indicator, titration of 20.0 cm³ of this solution required 18.7 cm³ of a hydrochloric acid solution for complete neutralisation. [atomic masses: Ba = 137, O = 16, H = 1)

• write the equation for the titration reaction.

Ba(OH)_2(aq) + 2HCl_(aq) ==> BaCl_2(aq) + 2H₂O_(l)

(b) Calculate the molarity of the barium hydroxide solution. 

formula mass of Ba(OH)₂ = 171, moles = 2.74 ÷ 171 = 0.016 mol in 100 cm³, (scaling up x 10)

therefore 0.16 mol in 1000 cm³, so molarity of Ba(OH)₂ = 0.16 mol dm^-3

(c) calculate the moles of barium hydroxide neutralised. 

moles Ba(OH)₂ used in titration = 0.16 x 20/1000 = 0.0032 mol

(d) calculate the moles of hydrochloric acid neutralised.

moles HCl titrated = 2 x moles of Ba(OH)₂ used (2 : 1 in equation) 2 x .0032

= 2 x 0.0032 = 0.0064 mol HCl in 18.7 cm³ of the acid solution, 18.7 cm³ = 0.0187 dm³

(e) calculate the molarity of the hydrochloric acid.

 therefore molarity of HCl_(aq) = 0.0064/0.0187 = 0.342 mol dm^-3

 

18. 90g of pure sulphuric acid was dissolved in water, the resulting total volume was 200 cm³. 20.7 cm³of this solution was found on titration, to completely neutralise 10.0 cm³ of a sodium hydroxide solution. [atomic masses: S = 32, O = 16, H = 1)

• Write the equation for the titration reaction.

2NaOH_(aq) + H₂SO_4(aq) ==> Na₂SO_4(aq) + 2H₂O_(l)

(b) calculate the molarity of the sulphuric acid solution.

moles H₂SO₄ = 4.90 ÷ 98 = 0.050 mol in 200cm³

scaling up to get molarity of the sulphuric acid solution, 0.050 x 1000 ÷ 200 = 0.25 mol dm^-3

(c) calculate the moles of sulphuric acid neutralised. 

 moles of sulphuric acid neutralised = 0.250 x 20.7/1000 = 0.005175 mol

(d) calculate the moles of sodium hydroxide neutralised. 

moles of sodium hydroxide neutralised = 2 x 0.005175

 = 0.01035 mol (2 : 1 in equation)

(e) calculate the concentration of the sodium hydroxide in mol dm^-3 (molarity).

concentration of the sodium hydroxide = 0.01035 x 1000 ÷ 10 = 1.035 mol dm-3 (molarity 1.04, 3sf)

 

19. 100 cm³of a magnesium hydroxide solution required 4.5 cm³ of sulphuric acid (of concentration 0.100 mol dm^-3) for complete neutralisation.

           [atomic masses: Mg = 24.3, O = 16, H = 1)

• Give the equation for the neutralisation reaction.

Mg(OH)_2(aq) + H₂SO_4(aq) ==> MgSO_4(aq) + 2H₂O_(l)

• Calculate the moles of sulphuric acid neutralised.

moles of sulphuric acid neutralised = 0.100 x 4.5/1000 

                                         = 0.00045 mol

• Calculate the moles of magnesium hydroxide neutralised.

moles of magnesium hydroxide neutralised also

                              = 0.00045 (1:1 in equation) in 100 cm³

 

• Calculate the concentration of the magnesium hydroxide in mol dm^-3(molarity).

concentration of the magnesium hydroxide

                            = 0.00045 x 1000 ÷ 100 = 0.0045 mol dm

(scaling up to 1000cm³ = 1dm³,  to get molarity)

 

(e) calculate the concentration of the magnesium hydroxide in g cm^-3.

                        molar mass of Mg(OH)₂ = 58.3

so concentration of the magnesium hydroxide 

                   = 0.0045 x 58.3 = 0.26 g dm^-3 (= g per 1000 cm³),

 

20. Magnesium oxide is not very soluble in water, and is difficult to titrate directly.Its purity can be determined by use of a ‘back titration’ method.

4.06 g of impure magnesium oxide was completely dissolved in 100 cm³ of hydrochloric acid, of concentration 2.00 mol dm^-3 (in excess).

The excess acid required 19.7 cm³ of sodium hydroxide (0.200 mol dm^-3) for neutralisation using phenolphthalein indicator and the end-point is the first permanent pink colour.

This 2nd titration is called a ‘back-titration’, and is used to determine the unreacted acid.

[atomic masses: Mg = 24.3, O = 16)

• (i) Why do you have to use excess acid and employ a back titration?

The magnesium oxide is insoluble in water and so cannot be titrated directly by a volumetric analysis method. By dissolving in excess acid you are sure to get all the MgO into solution, and then determine the excess acid, from which you can derive the amount of MgO that dissolved.

(ii) Write equations for the two neutralisation reactions.

MgO_(s) + 2HCl_(aq) ==> MgCl_2(aq) + H₂O_(l)

NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)

 

• Calculate the moles of hydrochloric acid added to the magnesium oxide.

moles of hydrochloric acid added to the magnesium oxide = 2 x 100/1000 = 0.20 mol HCl

• Calculate the moles of excess hydrochloric acid titrated.

moles of excess hydrochloric acid titrated

             = 19.7 ÷ 1000 x 0.200 = 0.00394 mol HCl

{mole ratio NaOH : HCl is 1 : 1 from equation (ii)}

• Calculate the moles of hydrochloric acid reacting with the magnesium oxide.

mole MgO reacted = 0.196 ÷ 2 

                   = 0.098 {1: 2 in equation (i)}

the formula mass of MgO = 40.3

therefore mass of MgO reacting with acid

                         = 0.098 x 40.3 = 3.95 g

• Calculate the moles and mass of magnesium oxide that reacted with the initial hydrochloric acid.

molar mass of Mg(OH)₂ = 58.3

so concentration of the magnesium hydroxide 

= 0.0045 x 58.3 = 0.26 g dm^-3 (= g per 1000 cm³),

so concentration = 0.26 ÷ 1000 = 0.00026 g cm^-3

• hence the % purity of the magnesium oxide.

% purity = 3.95 ÷ 4.06 x 100 

                = 97.3% MgO

21.    2.00 dm³of concentrated hydrochloric acid (10.0 M) was spilt onto a laboratory floor. It can be neutralised with limestone powder. [atomic masses: Ca = 40, C = 12, O = 16)

• give the equation for the reaction between limestone and hydrochloric acid.

 CaCO_3(s) + 2HCl_(aq) ==> CaCl_2(aq) + H₂O_(l) + CO_2(g)

• how many moles of hydrochloric acid was spilt?

moles of hydrochloric acid was spilt

                    = 2.00 x 10.0 = 20 mol HCl

• how many moles of calcium carbonate will neutralise the acid?

 moles of calcium carbonate to neutralise the acid

                      = 20 ÷ 2 = 10.0 mol CaCO₃ (1:2 in equation)

• what minimum mass of limestone powder is needed to neutralise the acid?

 formula mass of CaCO₃ = 100,

so mass of limestone powder needed to neutralise the acid

          = 100 x 10 = 1000g CaCO₃

 

(e) If 1000 dm³ of sulphuric acid, of concentration 2.00 mol dm^-3, leaked from a tank,

calculate the minimum mass of magnesium oxide required to neutralise it.  

the neutralisation reaction is MgO + H₂SO₄ ==> MgSO₄ + H₂O,

moles H₂SO₄ = 1000 x 2 = 2000 mol acid,

 2000 mol MgO needed (1:1 in equation),

mass MgO needed = 2000 x 40.3 = 80600 g or 80.6 kg

50. A 50.0 cm³sample of sulphuric acid was diluted to 1.00 dm³. A sample of the diluted sulphuric acid was analysed by titrating with aqueous sodium hydroxide. In the titration, 25.0 cm³ of 1.00 mol dm^-3 aqueous sodium hydroxide required 20.0 cm³ of the diluted sulphuric acid for neutralisation.

• Give the equation for the full neutralisation of sulphuric acid by sodium hydroxide.

         2NaOH_(aq) + H₂SO_4(aq) ==> Na₂SO_4(aq) + 2H₂O_(l)

(b) Calculate how many moles of sodium hydroxide were used in the titration?

moles of sodium hydroxide used in the titration

= 25.0 x 1/1000 = 0.025 mol NaOH

(c) Calculate the concentration of the diluted acid.

mol H₂SO₄ = mol NaOH ÷ 2 = 0.0125 mol in 20.0 cm³,

so scaling up to 1000 cm³ to get molarity of diluted acid

 = 0.0125 x 1000 ÷ 20 = 0.625 mol dm^-3

(or molarity = 0.0125 mol/0.02 dm³ = 0.625 mol dm^-3)

• Calculate the concentration of the original concentrated sulphuric acid solution.

scaling up from 50 to 1000 cm³, gives the concentration of the original concentrated sulphuric acid solution,

 = 0.625 x 1000 ÷ 50 = 12.5 mol dm^-3

23. A sample of sodium hydrogencarbonate was tested for purity using the following method. 0.400g of the solid was dissolved in 100 cm³of water and titrated with 0.200 mol dm^-3hydrochloric acid using methyl orange indicator. 23.75 cm³ of acid was required for complete neutralisation. [Ar’s: Na = 23, H = 1, C = 12, O = 16]

(a) Write the equation for the titration reaction. (a) NaHCO_3(s) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l) + CO_2(g)

(b) Calculate the moles of acid used in the titration and the moles of sodium hydrogencarbonate titrated.

 mol = molarity x volume in dm³, mol acid

         = 0.200 x 23.75/1000             

           = 4.75 x 10^-3 mol HCl

from equation HCl:NaHCO₃ is 1:1 by ratio

so mol HCl = mol NaHCO₃ = 4.75 x 10^-3

(c) Calculate the mass of sodium hydrogen carbonate titrated and hence the purity of the sample.

 mass = mol x formula mass, f. mass NaHCO₃ = 23 + 1 + 12 + (3 x 16) = 84

mass NaHCO₃ = 4.75 x 10^-3 x 84 

               =  0.399 g

% purity of NaHCO₃ = 0.399 x 100/0.40 

                            =  99.75% (99.8%, 3sf)

(d) If 0.400g of another group 1 hydrogen carbonate in its pure state, was titrated with the same acid and it took 20.00 cm3 to neutralise it, calculate …

(i) moles of acid needed for neutralisation and moles of hydrogen carbonate titrated)

mol = molarity x volume in dm³, mol acid

       = 0.200 x 20.00/1000

           = 4.00 x 10^-3 mol HCl

from equation above HCl : MHCO₃ is 1 : 1 by ratio

so mol HCl = mol MHCO₃ 

                        = 4.00 x 10^-3

(ii) the formula mass of the hydrogencarbonate

to get to the relative formula mass

moles = mass / M_r,  M_r = mass / mol

                            = 0.400 / 4.00 x 10^-3 = 100

(iii) by working out the atomic mass of M, suggest the identity of M in the group 1 hydrogencarbonate formula MHCO₃

 for MHCO₃, M_r(HCO₃) = 1 + 12 + 48 = 61, 

                          A_r (M) = 100 – 61 = 39,

which, from the periodic table relative atomic mass values, corresponds to potassium

So the formula of the group 1 hydrogencarbonate titrated was KHCO₃

 a)(i) Write out the equation, complete with state symbols for the reaction between hydrochloric acid and sodium carbonate. Na₂CO_3(aq) + 2HCl_(aq) ==> 2NaCl_(aq) + H₂O_(l) + CO_2(g)

Atomic masses: O = 16, H = 1, Na = 23, C = 12, Ca = 40, P = 31.09

(ii) A pipetted 25.0 cm³ aliquot of a solution of sodium carbonate is to be titrated with an approximately 1.0 mol dm^-3 hydrochloric acid to be standardised.

 20.0 cm³ of 1.0 mol dm^-3 hydrochloric acid contains 1.0 x 20.0/1000 = 0.02 mol HCl

From the equation, 0.020 mol HCl reacts with 0.010 mol Na₂CO₃,   M_r(Na₂CO₃) = 106

therefore mass Na₂CO₃ titrated = 0.01 x 106 = 1.06 g per aliquot,

since 250 cm³ is ¹/₁₀th of the aliquot,

          10 x 1.06 = 10.6 g of Na₂CO₃ would be used to make up the solution.

Molarity of Na₂CO_3(aq) = (10.6 g/106 g mol^-1)/0.25 dm³ 

                            = 0.40 mol dm^-3

 

24. What mass of dried anhydrous sodium carbonate must be dissolved in 250 cm³of deionised water, so that a 25.0 cm³aliquot of the carbonate solution will give a 20.0 cm³ titration with the hydrochloric acid?
25. What is the molarity of the sodium carbonate solution, assuming 100% purity.

(b)(i) The simplified molecular structure of 2-ethanoylhydroxybenzoic acid (‘Aspirin’) is CH₃COOC₆H₄COOH.

Give the equation of its reaction with sodium hydroxide.  CH₃COOC₆H₄COOH + NaOH ==> CH₃COOC₆H₄COO^–Na⁺ + H₂O

(ii) A sample of aspirin was to be analysed for purity by titrating it with standardised 0.100 mol dm^-3 sodium hydroxide using phenolphthalein indicator. Assuming 100% purity and access to a 4 decimal place electronic balance, calculate the mass of Aspirin that should be weighed out to give a titration of 23.0 cm³ of the alkali.

23.0 cm³ of 0.100 mol dm^-3 NaOH contains 0.100 x 23.0/1000 = 0.0023 mol NaOH

From the equation, mol Aspirin = mol NaOH, M_r(CH₃COOC₆H₄COOH) = 180

so need Aspirin mass of 0.0023 x 180 = 0.414 g

(iii) The main contaminant is likely to be unreacted 2-hydroxybenzoic acid. Why is this likely to be an impurity? and how will this affect the % purity you calculate i.e. why and how will the % purity be in error? 

The last stage in the synthesis of 2-ethanoylhydroxybenzoic acid (‘Aspirin’) is made by esterifying 2-hydroxybenzoic acid with ethanoic anhydride.

M_r(HOC₆H₄COOH) = 138, is 42 units less than aspirin. In terms of this particular impurity the % aspirin will be overestimated for the following reason. The 2-hydroxybenzoic acid will also be titrated with the aspirin, and, with its smaller molecular mass, it will need more alkali to neutralise than aspirin per equivalent mass of material. This can result in >100% purity!!!!

1. Outline complexiometric titration

Complexometric titrations are used mainly to determine metal ions by use of complex-forming reactions. Although many complexing agents (cyanide, thiocyanate, fluoride, 1,2-diaminoethane, etc.) can be used for this purpose, in practice the titrants are almost always compounds having the iminodiacetic acid functional groups.

The most widely applied are ethylenediaminetetraacetic acid, H₄Y and the dihydrate of the sodium salt, Na₂H₂Y ⋅ 2H₂O (better soluble in water).

                            Ethylenediaminetetraacetic acid, H4Y

1. Explain the properties of EDTA that makes it suitable for complexiometric titration

EDTA (ethylene diamine tetraacetic acid) possesses several properties that make it highly suitable for complexometric titrations. These properties contribute to the effectiveness and reliability of EDTA as a complexing agent in these types of titrations. The key properties of EDTA include:

• Chelating Ability: One of the essential properties of EDTA is its ability to form stable chelate complexes with metal ions. EDTA contains four carboxylic acid groups and two amine groups, which allow it to form strong bonds with metal ions through coordination chemistry. The formation of chelate complexes enhances the stability and specificity of the complex, making EDTA highly selective for metal cations.
• High Affinity for Metal Ions: EDTA has a high affinity for metal ions due to the presence of multiple donor atoms in its structure. The four oxygen atoms from the carboxylic acid groups and the two nitrogen atoms from the amine groups can donate electron pairs to form coordinate bonds with metal ions. This high affinity ensures the efficient complexation of metal ions and contributes to the accuracy of complexometric titrations.
• pH Dependence: The complexation reaction between EDTA and metal ions is pH-dependent, but within a specific pH range (usually around pH 2-10), the reaction is relatively insensitive to pH changes. This property allows for complexometric titrations to be carried out under a wide range of pH conditions, providing flexibility in experimental setups.
• Solubility: EDTA is highly soluble in water, making it easy to prepare standardized solutions of known concentration. This solubility facilitates the accurate and precise measurement of EDTA volumes required for titrations, ensuring reliable results.
• Stoichiometry: The complexation reaction between EDTA and metal ions generally follows a 1:1 stoichiometry, meaning one metal ion reacts with one EDTA molecule to form a complex. This stoichiometric ratio simplifies the determination of metal ion concentrations by measuring the volume of EDTA solution required for complete complex formation.
• Stability: The chelate complexes formed by EDTA with metal ions are highly stable. The stability constants of these complexes are generally large, ensuring that the complexes remain intact during the course of titration. This stability allows for precise determination of metal ion concentrations and enhances the accuracy and reliability of complexometric titrations.
• Versatility: EDTA can form complexes with a wide range of metal ions, including divalent and trivalent metal cations. This versatility enables EDTA to be used for the determination of various metal ions in different samples, making it a valuable tool in analytical chemistry.

These properties collectively make EDTA well-suited for complexometric titrations, providing reliable and accurate determination of metal ion concentrations in complex samples.

List advantages of EDTA  as a standard in complexiometric titration

EDTA (ethylene diamine tetraacetic acid) is a widely used complexing agent in complexometric titration due to its unique properties. Some advantages of using EDTA as a standard in complexometric titration are:

• High Selectivity: EDTA forms stable complexes with a wide range of metal ions, making it highly selective for metal cations. It can complex with divalent and trivalent metal ions, including calcium, magnesium, zinc, copper, iron, and many others. This selectivity allows for precise and accurate determination of metal ions in complex samples.
• Chelation Effect: EDTA acts as a chelating agent, meaning it can form multiple bonds with a metal ion simultaneously. It forms a six-membered ring structure with the metal ion, known as a chelate complex. This chelation effect increases the stability and strength of the complex, enhancing the accuracy of complexometric titrations.
• pH Independence: The complexation reaction of EDTA with metal ions is pH-independent within a certain pH range (usually around pH 2-10). This pH independence allows for the titration to be conducted under a wide range of pH conditions without affecting the accuracy of the results.
• Quantitative Complex Formation: EDTA forms 1:1 stoichiometric complexes with most metal ions, resulting in a simple and well-defined reaction equation. This allows for precise determination of the metal ion concentration by measuring the volume of EDTA solution required for complete complex formation.
• Versatility: EDTA can be used for the determination of a wide range of metal ions, making it versatile for various applications. It finds applications in environmental analysis, pharmaceutical analysis, water treatment, food analysis, and many other fields.
• Stability and Solubility: EDTA and its metal-EDTA complexes are stable under normal laboratory conditions, which ensures reliable and reproducible results. Additionally, EDTA is highly soluble in water, making it easy to prepare standardized solutions and to work with in titration procedures.
• Availability and Cost-Effectiveness: EDTA is commercially available and relatively inexpensive compared to other complexing agents. Its widespread use and availability make it a cost-effective option for complexometric titrations.

These advantages highlight the utility of EDTA as a standard in complexometric titrations, enabling accurate and reliable determination of metal ion concentrations in various samples.

1. State properties of Eriochrome Black T(EBT) that makes it suitable indicator in Ethylenediamine tetra acetic acid (EDTA) titration.

Eriochrome Black T is a complexometric indicator that is used in complexometric titrations, e.g. in the water hardness determination process. It is an azo dye. In its deprotonated form, Eriochrome Black T is blue. It turns or changes  colour from blue to pink in the process when it forms a complex with calcium, magnesium, or other metal ions.

The dye–metal ion complex is less stable than the EDTA–metal ion complex.

Eriochrome black T is used as an indicator for complexometric titrations because it forms a complex with calcium, magnesium and other metal ions in its protonated form. When titrated with EDTA, the metal ions complexed with eriochrome black T reacts with the EDTA forming a blue solution.

100. Outline the process of preparing 0.1M Silver nitrate solution given deionized water , clean and dry volumetric flask (1000ml) and silver nitrate crystals
     1. Pure calcium carbonate can be used to make a standard calcium ion solution to practice a complexometric titration of calcium ions with EDTA or determine the molarity of the EDTA reagent.

• Give a simple equation to show the chelation reaction between hydrated calcium ions and the EDTA anion at pH10 and what sort of reaction is it?

 It is a ligand substitution/replacement reaction.

[Ca(H₂O)₆]²⁺_(aq) + EDTA^4-_(aq) ==> [CaEDTA]^2-_(aq) + 6H₂O_(l)

more simply Ca²⁺_(aq) + EDTA^4-_(aq) ==> [CaEDTA]^2-_(aq)

or [Ca(H₂O)₆]²⁺_(aq) + H₂EDTA^2-_(aq) ==> [CaEDTA]^2-_(aq) + 2H⁺_(aq) + 6H₂O_(l)

more simply Ca²⁺_(aq) + H₂EDTA^2-_(aq) ==> [CaEDTA]^2-_(aq) + 2H⁺_(aq)

• To make a standard calcium ion solution 0.250 g of A.R. calcium carbonate was dissolved in a little dilute hydrochloric acid and made up to 250 cm³in a calibrated volumetric flask. Calculate the molarity of the calcium ion in this solution.

 M_r(CaCO₃) = 100, mol CaCO₃ = mol Ca²⁺ in solution

                                              = 0.250/100 = 0.00250 mol

since 250 cm³ = 0.25 dm³, molarity Ca²⁺ = 0.0025/0.25 

                                                                = 0.010 mol dm^-3

1. Approximately 1.0g of the solid disodium dihydrate salt of EDTA was dissolved in 250 cm³of water in a volumetric flask. 25.0 cm³ of this was pipetted into a conical flask and ~1 cm³ of a conc. ammonia/ammonium chloride pH10 buffer was added. After adding a few drops of Eriochrome Black T indicator, the EDTA solution was titrated with the standard calcium ion solution (from part ii) until the reddish tinge turns to blue at the endpoint. If 25.7 cm³ of the EDTA solution was required to reach the equivalence point, what was the molarity of the EDTA?

mole CaCO₃ = mol Ca²⁺ = mol EDTA used in titration.

Therefore from c(ii) mol Ca²⁺ = mol EDTA

                                    = 0.01 x 25.0/1000

              = 0.00025 mol in 25.70 cm³ (0.0257 dm³) EDTA solution,

so molarity EDTA = 0.00025/0.0257 

                          = 0.00973 mol dm^-3 (3 sf)

1. In human teeth, approximately 96% of the outer enamel and 70% of the inner dentine are composed of the apatite mineral, calcium hydroxy phosphate. If the simplest empirical formula is Ca₅(PO₄)₃OH

calculate the % calcium in the apatite mineral.

M_r(apatite) = (5 x 40) + 3 x (31 + 4 x 16) + (16 + 1) = 502

% Ca in apatite = 200 x 100/502

                           = 39.8%

1. A dried 1.40g human tooth was dissolved in a small quantity of hot conc. nitric acid. A drop of methyl orange indicator was added followed by drops of 6M sodium hydroxide until the indicator turned orange to neutralise the solution. The solution was then made up to 250 cm³in a volumetric flask. 10.0 cm³ of this solution was pipetted into a conical flask and ~1 cm³ of a conc. ammonia/ammonium chloride pH10 buffer was added.

This solution was then titrated with 0.0200 mol dm^-3 EDTA using Eriochrome Black T indicator. The indicator turned blue after 22.5 cm³ of EDTA was added. Calculate the average % by mass of calcium throughout the tooth.

 In the titration mol Ca²⁺ = mol EDTA,

therefore mol Ca²⁺ = 22.5 x 0.0200/1000 = 0.00045,

since ¹⁰/₂₅₀ of the original solution was used in the titration,

the total mol of calcium in the tooth solution = 0.00045 x 250/10

                                                                     = 0.01125 mol Ca

total mass of Ca in tooth = 0.01125 x 40

                                        =  0.45 g

% by mass Ca in the tooth = 0.45 x 100/1.40

                                         = 32.1 %

25. 0 cm³of seawater was diluted to 250 cm³ in a graduated volumetric flask. 

A 25.0 cm³ aliquot of the diluted seawater was pipetted into a conical flask and a few drops of potassium chromate(VI) indicator solution was added.

 On titration with 0.100 mol dm^-3 silver nitrate solution, 13.8 cm³ was required to precipitate all the chloride ion.

[Atomic masses: Na = 23, Cl = 35.5]

• Give the ionic equation for the reaction of silver nitrate and chloride ion.

 Ag⁺_(aq) + Cl^–_(aq) ==> AgCl_(s) (sodium ions and nitrate ions etc. are spectator ions)

• Calculate the moles of chloride ion in the titrated 25.0 cm³

from equation: moles silver nitrate (AgNO₃)  = moles chloride ion (Cl)

moles = molarity AgNO₃ x volume of AgNO₃ in dm³

= 0.100 x 13.8/1000

  = 1.38 x 10^-3 mol Cl^– (in 25.0 cm³ aliquot)

• Calculate the molarity of chloride ion in the diluted seawater.

moles in 1 dm³ of diluted seawater = 1.38 x 10^-3 x 1000/25

                           = 0.0552 (scaling up to 1000 cm³)

So molarity of chloride in diluted seawater is 0.0552 mol dm^-3

• Calculate the molarity of chloride ion in the original seawater.

Now in the titration 25.0 cm³ of the 250 cm³ was used,

so the molarity of chloride ion in the original seawater must be scaled up accordingly.

molarity of chloride ion in seawater = 0.0552 x 250/25.0 

                                                          = 0.552 mol dm^-3

(e) Assuming that for every chloride ion there is a sodium ion, what is the theoretical concentration of sodium chloride salt in g dm^-3 in seawater?

 M_r(NaCl) = 23 + 35.5 = 58.5

concentration of NaCl in g dm^-3 = molarity x formula mass

                        = 0.552 x 58.5 =  32.3 g dm^-3

33. 0.12 g of rock salt was dissolved in water and titrated with 0.100 mol dm^-3silver nitrate until the first permanent brown precipitate of silver chromate is seen.

19.7 cm³ was required to titrate all the chloride ion.

[Atomic masses: Na = 23, Cl = 35.5]

• How many moles of chloride ion was titrated?

moles = molarity AgNO₃ x volume in dm³ 

= 0.100 x 19.7/1000

 = 1.97 x 10^-3 mol Cl^– ion

[AgNO₃:NaCl or Ag⁺ : Cl^– is 1 : 1,

mass NaCl = 23 + 35.5 = 58.5

• What mass of sodium chloride was titrated?

mass = mol x formula mass = 1.97 x 10^-3 x 58.5

 = 0.1152 g NaCl

• What was the % purity of the rock salt in terms of sodium chloride?

% purity = 0.1152 x 100/0.12 

= 96.0 % in terms of NaCl (3sf)

5. 00g of a solid mixture of anhydrous calcium chloride(CaCl₂) and sodium nitrate (NaNO₃) was dissolved in 250 cm³ of deionised water in a graduated volumetric flask. A 25.0 cm³ aliquot of the solution was pipetted into a conical flask and a few drops of potassium chromate(VI) indicator solution was added.

On titration with 0.1 mol dm^-3 silver nitrate solution, until the first permanent brown precipitate of silver chromate was formed, 21.2 cm³ was required to precipitate all the chloride ion. [Atomic masses: Ca = 40, Cl = 35.5]

• Calculate the moles of chloride ion titrated.

mole Cl^– = moles Ag⁺ [=AgNO₃, see Q10(a)/(b)]

mole Cl^– = molarity AgNO₃ x vol AgNO₃ 

= 0.100 x 21.2/1000

 = 2.12 x 10^-3 mol Cl^–

(b) Calculate the equivalent moles of calcium chloride titrated.

Since calcium chloride is CaCl₂, mol CaCl₂ = mole Cl^–/2 

          = 2.12 x 10^-3/2 

              = 1.06 x 10^-3 mol CaCl₂

• Calculate the equivalent mass of calcium chloride titrated.

 M_r(CaCl₂) = 40 + (2 x 35.5) = 111

mass = mol x f. mass

= 1.06 x 10^-3 x 111 

             = 0.1177 g CaCl₂

(d) Calculate the total mass of calcium chloride in the original 5.0 g of the mixture. Since ¹/₁₀th of original solution titrated, original mass of CaCl₂ in mixture

= 10 x 0.1177 g 

 = 1.177g CaCl₂ (1.78g 3sf)

• The % of calcium chloride and sodium nitrate in the original mixture.

Therefore % = 1.177 x 100/5.0 

=  23.5% CaCl₂ (3 sf)

and % NaNO₃ = 100 – 23.5  

      = 76.5% (3 sf)

34.  A bulk solution of hydrochloric acid was standardised using pure anhydrous sodium carbonate (Na₂CO₃, a primary standard).

13.25 g of sodium carbonate was dissolved in about 150.0 cm³ of deionised water in a beaker.

The solution was then transferred, with appropriate washings, into a graduated flask, and the volume of water made up to 250 cm³, and thoroughly shaken (with stopper on!) to ensure complete mixing.

25.0 cm³ of the sodium carbonate solution was pipetted into a conical flask and screened methyl orange indicator added. The aliquot required 24.65 cm³ of a hydrochloric acid solution, of unknown molarity, to completely neutralise it. [atomic masses: Na = 23, C = 12, O = 16]

(a) Calculate the molarity of the prepared sodium carbonate solution.

 moles = mass/f. mass, f. mass Na₂CO₃ = 106, mol

 Na₂CO₃ = 13.25/106 = 0.125

molarity = mol/vol. in dm³, 250 cm³  = 0.250 dm³,

molarity Na₂CO₃ = 0.125/0.250 

= 0.50 mol dm^-3

(b) Write out the equation between sodium carbonate and hydrochloric acid, including state symbols. 

Na₂CO_3(aq) + 2HCl_(aq) ==> 2NaCl_(aq) + H₂O_(l) + CO_2(g)

(c) How many moles of sodium carbonate were titrated?

mol = molarity x volume

mol Na₂CO₃ titrated = 0.5 x 25.0/1000 = 0.0125 mol Na₂CO₃ (in the 25 cm³ aliquot pipetted)

• How many moles of hydrochloric acid were used in the titration?

 from equation, mole ratio Na₂CO₃:HCl is 1:2,

so mol HCl = 2 x mol Na₂CO₃  = 2 x 0.0125 

= 0.025 mol HCl (in the 24.65 cm³ titre)

(e) What is the molarity of the hydrochloric acid?

 molarity = mol/vol. in dm³,

         dm³ = cm³/1000, 24.65/1000 = 0.02465 dm³

therefore molarity HCl = 0.025/0.02465 

                           = 1.014 mol dm^-3 (1.01 3sf)

36. Describe a procedure that can used to determine the molecular mass of an organic acid by titration with standardised sodium hydroxide solution. Indicate any points of the procedure that help obtain an accurate result and explain your choice of indicator.

An appropriate quantity of the acid is weighed out, preferably on a 4 sf electronic balance. It can be weighed into a conical flask by difference i.e. weight acid added to flask = (weight of boat + acid) – (weight of boat).  The acid is dissolved in water, or aqueous-ethanol if not very soluble in water. The solution is titrated with standard sodium hydroxide solution using phenolphthalein indicator until the first permanent pink.

The burette should be rinsed with the sodium hydroxide solution first. During the titration the flask should be rinsed around the inside to ensure all the acid and alkali react, and drop-wise addition close to the end-point to get it to the nearest drop – the first permanent pink colour.

37. 0.279g of an organic monobasic aromatic carboxylic acid, containing only the elements C, H and O, was dissolved in aqueous ethanol. A few drops of phenolphthalein indicator were added and the mixture titrated with 0.100 mol dm^-3sodium hydroxide solution.

It took 20.5 cm³ of the alkali to obtain the first permanent pink. [at. masses: C = 12, H = 1 and O = 16]

The pK_ind for phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a solution of the sodium salt of the acid (from strong base + weak acid) is in this region and so the equivalence point can be detected with this indicator.

(b) How many moles of sodium hydroxide were used in the titration? moles = molarity x volume in dm³ (dm³ = cm³/1000)

mol NaOH = 0.100 x 20.5/1000 

= 0.00205 mol

(c) How many moles of the organic acid were titrated? and explain your reasoning.

mol NaOH = mol RCOOH

                 = 0.00205

because 1:1 mole ratio for a monobasic acid: RCOOH + Na⁺OH^– ==> RCOO^–Na⁺ + H₂O

• Calculate the molecular mass of the acid. moles = mass (g)/M_r, so M_r= mass/mol = 0.279/0.00205

                                         = 136.1

(e) Suggest possible structures of the acid with your reasoning.

 The simplest aromatic carboxylic acid is benzoic acid C₆H₅COOH, M_r = 122

136-122 = 14, which suggests an ‘extra’ CH₂ (i.e. -CH₃ attached to the benzene ring instead of a H), so, since the COOH is attached to the ring, there are three possible positional/chain isomers of CH₃C₆H₄COOH (M_r = 136)

2-, 3- or 4-methylbenzoic acid.

 

38.  0.103g of a dibasic/diprotic non-aromatic carboxylic acid required 19.85 cm³of a standardised sodium hydroxide solution for complete neutralisation. If the concentration of the alkali was 0.0995 mol dm^-3. [at. masses: C = 12, H = 1 and O = 16]

Calculate …

(a) moles of sodium hydroxide used in the titration, moles = molarity x volume in dm³ (dm³ = cm³/1000)

mol NaOH = 0.0995 x 19.85/1000 

                = 0.001975 mol

(b) Moles of dibasic/diprotic acid titrated giving your reasoning, The titration reaction for complete neutralisation is:

R(COOH)₂ + 2Na⁺OH^– ==> R(COO^–Na⁺)₂ + 2H₂O

this 1 : 2 reaction mole ratio means that mol dibasic acid = mol NaOH/2

= 0.001975/2 = 0.0009875

• the molecular mass of the acid, moles = mass (g)/M_r,

so M_r = mass/mol = 0.103/0.0009875 = 104.3 (approx. 104 3sf)

(d) a possible structure of the acid.

 Since a dibasic acid, and 2 x COOH = 2 x 45 = 90 mass units, the remaining 14 units could be CH₂, and so the structure is likely to be HOOC-CH₂-COOH, propanedioic acid (malonic acid), M_r = 144

 

339. 0.236g of benzoic acid required 19.25 cm³of 0.100mol dm^-3sodium hydroxide for complete neutralisation.

Calculate …

• moles of sodium hydroxide used in titration,

                  mol NaOH = 0.100 x 19.25/1000

                                 = 0.001925

(b) Moles and mass of benzoic acid titrated [at. masses: C = 12, H = 1 and O = 16]

mol NaOH = mol acid = 0.001925, mol ratio 1:1,

C₆H₅COOH + Na⁺OH^– ==> C₆H₅COO^–Na⁺ + H₂O

M_r (C₆H₅COOH) =122, mol = mass (g)/M_r or mass = mol x M_r

so mass acid = 0.001925 x 122 

= 0.2349 g

(c) % purity of benzoic acid from this assay titration.

 % purity = actual mass of acid titrated x 100 / mass of original sample   % purity = 0.2349 x 100/0.236 = 99.5% (3 sf)

40. Sodium hydroxide solution can be standardised. 0.250 g of very pure benzoic acid (C₆H₅COOH) was titrated with a solution of sodium hydroxide of unknown molarity. If 22.5 cm³of the alkali was required for neutralisation, Calculate …

• moles of acid titrated [at. masses: C = 12, H = 1 and O = 16],

M_r (C₆H₅COOH) =122, mol acid = mass (g)/M_r 

                            = 0.250/122 

                            = 0.002049 mol

• mol alkali used in titration,

mol alkali = 0.002049 mol, since mol acid, 1:1 mole ratio in reaction (see Q17(b)).

(c) the molarity of the alkali.

 Since 0.002049 mol NaOH in 22.5 cm³, so scaling up

molarity NaOH = 0.002049 x 1000/22.5

                     =  0.0911 mol dm^-3 (3 sf)

41. The solubility of calcium hydroxide in water can be measured reasonably accurately to 3sf by titrating the saturated solution with standard hydrochloric acid.

(a) If the standard hydrochloric acid is made by diluting ‘2M’ bench acid, what volume of the ‘2M’ acid is required to make up 250 or 500 cm³ of approximately 0.1 mol dm^-3 hydrochloric acid and how might you do it? The ratio of 2M/0.1M is 20, so need to do 20 fold dilution.

So 25.0 cm³ of 2M diluted to 500 cm³ gives an approximately 0.1 mol dm^-3 solution. You can do this with a measuring cylinder and beaker.

In practice you could make up 12.5 cm³ of the 2M acid up to 250 cm³. You could do this with a burette and a 250 cm³ standard volumetric flask, but standardisation of the acid is still required.

• Why must the 2M acid be diluted and why must the diluted acid be standardised?

The solubility of calcium hydroxide is low, so it would give a very inaccurate tiny titration value with relatively concentrated acid. For any accurate work e.g. to 3sf, standardisation of reagents is required.

In the calculation below assume the molarity of the standardised hydrochloric acid is 0.1005 mol dm^-3.

At 25^oC, a few grams of solid calcium hydroxide was shaken with about 400 cm³ of deionised water, and then filtered. 50.0 cm³ samples of the ‘limewater’ gave an average titration of 15.22 cm³ of 0.1005 mol dm^-3 hydrochloric acid using phenolphthalein indicator. If the acid is in the burette, how would you measure out the calcium hydroxide solution? and why is phenolphthalein indicator used?

 1 x 50.0 cm³ pipette or 2 x 25.0 cm³ pipette would be the most convenient (or accurate burette?). Phenolphthalein is used for strong base-strong acid titrations.

• Give the equation for calcium hydroxide reacting with hydrochloric acid.

 Ca(OH)_2(aq) + 2HCl_(aq) ==> CaCl_2(aq) + 2H₂O_(l)

• What is the reacting mole ratio of Ca(OH)₂: HCl and hence calculate the moles of them involved in the titration.

from equation, mole ratio Ca(OH)₂ : HCl is 1:2

and since moles solute = molarity x volume in dm³

mol HCl used in titration = 0.1005 x 15.22/1000

                                 = 0.001530 mol HCl

therefore mol Ca(OH)₂ = 0.001558/2 

                              = 0.000765 mol Ca(OH)₂

• Calculate the molarity of the solution in terms of mol Ca(OH)₂dm^-3.

Mr[Ca(OH)₂] =74, so solubility in g dm^-3 

          = 0.153 x 74 

          = 1.13 g dm^-3 (3 sf)

• What is the approximate solubility of calcium hydroxide in g Ca(OH)₂per 100g water?

 Since density of water is ~1.0 g cm^-3, the solubility of Ca(OH)₂ is about 0.113 g/100 cm³ H₂O