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ACIDS, BASES BUFFERS AND PH ANALYSIS
- List two properties of an acid and bases
An acid is basically a molecule which can donate an H+ ion and can remain energetically favourable after a loss of H+.
An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition).. :
- Aqueous solutions of acids are electrolytes, meaning that they conduct electrical current. Some acids are strong electrolytes because they ionize completely in water, yielding a great many ions. Other acids are weak electrolytes that exist primarily in a non-ionized form when dissolved in water.
- Acids have a sour taste. Lemons, vinegar, and sour candies all contain acids.
- Acids change the color of certain acid-base indicates. Two common indicators are litmus and phenolphthalein. Blue litmus turns red in the presence of an acid, while phenolphthalein turns colorless.
- Acids react with active metals to yield hydrogen gas. Recall that an activity series is a list of metals in descending order of reactivity. Metals that are above hydrogen in the activity series will replace the hydrogen from an acid in a single-replacement reaction, as shown below:
- Zn(s)+H2SO4(aq)→ZnSO4(aq)+H2(g)
Acids react with bases to produce a salt compound and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The products of this reaction are an ionic compound, which is labeled as a salt, and water.
Examples: Sulfuric acid [H2SO4], Hydrochloric acid [HCl], Acetic acid [CH3COOH].
Bases, on the other hand, are characterized by a bitter taste and a slippery texture. A base that can be dissolved in water is referred to as an alkali. When these substances chemically react with acids, they yield salts. Bases are known to turn red litmus blue. A base is a substance that can accept protons or donate a pair of valence electrons to form a bond
- Bases have properties that mostly contrast with those of acids.
- Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.
- Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.
- Bases also change the color of indicators. Litmus turns blue in the presence of a base while phenolphthalein turns pink.
- Bases do not react with metals in the way that acids do.
- Bases react with acids to produce a salt and water.
- Examples: Sodium hydroxide [NaOH], milk of magnesia [Mg(OH)2], calcium hydroxide [Ca(OH)2].
- Define an acid and base according to the following theories
- Lewis
- Bronsted -Lowry
- Distinguish between ;
a strong and a weak acid and bases
A strong acid or alkali is one that is nearly or completely ionised in water. Examples are: hydrochloric acid, nitric acid and sulphuric acid. Sodium and potassium hydroxide are examples of strong bases.
A weak acid or alkali, on the other hand, is only partially ionised in water. Ethanoic, citric and carbonic acids are all weak acids. The ethanoic acid molecule and the ions reach a dynamic equilibrium with the equilibrium normally well to the left, so there is little H+ present.
(b) Concentrated and dilute acid and bases
The concentration of an acid or alkali is a measure of how much acid or alkali is dissolved in a known volume of liquid, e.g. water. The molarity of a solution is a measure of concentration: the number of moles dissolved in 1000cm3 of water. Concentration of solutions is usually expressed in moles per litre of solution. (A litre is the same as 1dm3 or 1000cm3.). This way of stating concentration is sometimes called ‘molarity’.
In a concentrated acid or alkali there is a relatively large amount of solute dissolved, compared with a dilute acid or alkali. However, this does not tell us what the concentration of H+ ions is in the acid or OH- ions is in the alkali. This depends on the strength of the acid or alkali. Strength refers to the degree of ionisation in water.
4. Define PH
pH is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution.
Its is determined as pH = -log[H+] Where log is the base-10 logarithm and [H+] stands for the hydrogen ion concentration in units of moles per liter solution. The pH scale usually ranges from 0 to 14. Aqueous solutions at 25°C with a pH less than 7 are acidic, while those with a pH greater than 7 are basic or alkaline. A pH level of 7.0 at 25°C is defined as “neutral” because the concentration of H3O+ equals the concentration of OH− in pure water. Very strong acids might have a negative pH, while very strong bases might have a pH greater than 14.
5. Explain the meaning of the term
- Indicator
chemical indicator, any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution.
- universal indicator
A universal indicator is a mixture of different types of indicators that exhibits different coloration at different levels. It is used to detect the acidic or basic nature of a substance or a solution. It can be in the form of a paper strip or a solution. Example: Methyl red, and Phenolphthalein.
6. Distinguish between an amphoteric oxide and an acidic oxide giving an example in each case
An acidic oxide is an oxide that when combined with water gives off an acid. A basic oxide is an oxide that when combined with water gives off a base. When a substance reacts chemically, both as a base or acid is termed as amphoteric oxide
7. HCl is a strong acid and is completely dissociated in its solutions according to the process:
HCl(aq)→ H+(aq) + Cl–(aq)
From this process it is clear that one mole of HCl would give one mole of H+ ions. Therefore, the concentration of H+ ions would be equal to that of HCl i.e. 0.001 molar or 1.0 × 10–3 mol L–1.
Thus, [H+] = 1 × 10–3 mol L–1
pH = –log[H+] = –(log 10–3)
= –(–3 × log10) = –(3 × 1) = 3
Thus, pH = 3
9. What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10–5mol L–1 in concentration.
Sulphuric acid dissociates in water as:
H2SO4(aq)→ 2H+(aq) + SO42–(aq)
Each mole of sulphuric acid gives two mole of H+ ions in the solution. One litre of 5 × 10–5 mol L–1 solution contains 5 × 10–5 moles of H2SO4 which would give 2 × 5 × 10–5 = 10 × 10–5 or 1.0 × 10–4 moles of H+ ion in one litre solution.
Therefore,
[H+] = 1.0 × 10–4 mol L–1
pH = –log[H+] = –log10–4 = –(–4 × log10)
=–(–4×1)=4
Calculate the pH of 1 × 10–4molar solution of NaOH.
NaOH is a strong base and dissociate in its solution as:
NaOH(aq) → Na+(aq) + OH–(aq)
One mole of NaOH would give one mole of OH– ions. Therefore,
[OH–] = 1 × 10–4 mol L–1
pOH = –log[OH–] = –log × 10–4 = – (–4)
= 4
Since pH + pOH = 14
pH = 14 – pOH = 14 – 4
= 10
11. Calculate the PH of 0.1 M NaOH solution
0.1M. NaOH is a strong base, so this will produce 0.1mol/L of OH ions in solution. This will produce a pH of 13.
pOH = -log [1×10−1]
=−log1+(−log10−1)
= 0 + ( – ( – 1)) log 10
= 0 + 1 x 1
=1
Now we know that,
pH +pOH = 14
pH + 1 = 14
pH = 14 – 1
pH = 13
12. Calculate the pH of a solution in which the concentration of hydrogen ions is 1.0 × 10–8mol L–1.
Here, although the solution is extremely dilute, the concentration given is not of acid or base but that of H+ ions. Hence, the pH can be calculated from the relation:
pH = –log[H+]
given [H+] = 1.0 × 10–8 mol L–1
pH = –log10–8 = –(–8 × log10)
=–(–8×1) =8
Consider the mixture is made from a monobasic weak acid HA and an alkali metal salt M+A–
e.g. A = CH3COO, M = K or Na
it is reasonable to assume for simple approximate calculations that ..
[A–(aq)] = [salt(aq)] since salt fully ionised and M+ is a spectator ion, and
[HA(aq)]equilib., = [HA(aq)]initial since little of the weak acid is ionised.
Therefore the weak acid Ka expression is …
|
(i) Ka = |
[H+(aq)] [A–(aq)] |
|
––––––– mol dm–3 |
|
|
[HA(aq)] |
becomes
|
(ii) Ka = |
[H+(aq)] [salt(aq)] |
|
––––––––– mol dm–3 |
|
|
[acid(aq)] |
therefore: [H+(aq)] = Ka [acid(aq)] mol dm–3
[salt(aq)] ,
and taking –log10 of both sides gives
(iii) pHbuffer = –log10(Ka x [acid(aq)]
[salt(aq)])
or (iv) pHbuffer = pKa + –log10([acid(aq)]
[salt(aq)])
or (v) pHbuffer = pKa + log10([salt(aq)]
[acid(aq)])
which is how the equation is usually quoted, sometimes called the Henderson Equation.
Note: For a given conjugate pair (HA and A–), the pH of the buffer is determined by the acid/salt ratio, though the more concentrated the buffer, the greater its capacity to neutralise larger amounts of added/formed in a reaction medium.
Another point is how to choose which weak acid is best for a desired buffer?
The useful range of a buffer is decided by the weak acid’s Ka and the ratio of the salt and weak acid concentrations.
The buffer will be most useful when the ratio [salt]/[acid] is equal to one i.e. when both active ingredients are at their maximum concentrations at no expense to the other – by the principles of related chemical equilibrium, if you increase one concentration you would decrease the other.
Therefore the maximum buffer capacity is when [salt] = [acid]
Now Ka = [H+(aq)] when [salt] = [acid] in equation (i) or (ii) above
therefore taking –log10 of both sides gives …
pKa = pH when [salt] = [acid], because log10(1) = 0 in equations (iv) or (v) and this simple mathematical argument gives the necessary guidance …
So for example, supposing you wanted a buffer to cover a pH range of 4.5 to 6.0,and your choice of weak acid pKa‘s was 2.8, 4.2, 5.5 and 6.5,
you would choose the weak acid with a pKa of 5.5 because its pKa is well in the desired pH range.
You would then formulate it with its sodium (or potassium) salt – note that sodium ion and potassium ion salts are usually used because they have virtually no acidic or basic character to complicate matters e.g. the hydrated ions Na+(aq) and K+(aq) do not donate protons in the way, for example, the hexa–aqua ions of aluminium can …
[Al(H2O)6]2+(aq) + H2O(l) [Al(H2O)5(OH)]+(aq) + H3O+(aq)
because the polarising power of the central metal ions, Na+ or K+ is too small to effect this process.
14. Define
- Buffer solution
Buffers are a type of solution that resist a change in pH when an acid or base is added to it. Buffer solutions usually contain a weak acid or base combined with its conjugate acid or base.
- Buffering capacity
Buffer capacity is defined as the number of moles of acid or base that have to be added to 1 liter to cause its pH to change by 1 unit.
Buffer capacity (β) is defined as the moles of an acid or base necessary to change the pH of a solution by 1, divided by the pH change and the volume of buffer in liters; it is a unitless number. A buffer resists changes in pH due to the addition of an acid or base though consumption of the buffer.
15. Explain
- uses of buffer solution
Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. In nature, there are many living systems that use buffering for pH regulation. For example, the bicarbonate buffering system is used to regulate the pH of blood, and bicarbonate also acts as a buffer in the ocean.
- Working principles of buffer solution
Buffers follow the Le-Chatelier principle. It states that “the position of equilibrium will move in such a way as to counteract the change”. This means that, upon adding either a base or an acid to a buffer solution, equilibrium will behave in such a way, to minimize the effect of added acid or base as much as possible.
For example
If sodium acetate (CH3COONa) is added to acetic acid (CH3COOH) solution, it does not react directly with acetic acid or with water. However, the increase of acetate ions from (CH3COONa) suppresses the ionization of acetic acid.
The following reactions sum up to show a common ion, that is acetate ion (CH3COO–);
The mixture of CH3COOH and CH3COONa, in specific molar ratios in water, makes a buffer solution. The acidity of this solution is less than an acetic acid solution because the acetate ions hinder the ionization of acetic acid.
Now, consider a buffered solution containing a relatively large amount of weak acid HA and its conjugate base A–. The base is obtained from its salt i.e. Na-A. When hydroxide (OH–) ions are added to the solution, they react with acid (HA) readily to form water, a neutral molecule. This is how a change in pH of the solution is resisted by the buffer solutions, upon the addition of a base.
17. Describe how you would prepare a phosphate buffer with a pH of about 7.40
For a buffer to function correctly , the concentration of the acid component must be roughly equal to the conjugate base component.
When the desired pH is close to the pKa of the acid , i.e. when pH=pKa
log (conjugate base) =0
(acid) (conjugate base) =1
(acid)
Because phosphoric acid is a trprotic acid , it undergoes three stages of ionization i.e. ;
H3PO4H+ ⇌ +H2PO4– Ka=7.5X10-3 Pka=2.12
H2POH+ ⇌ +HPO4-2 Ka=6.2X10-8Pka=7.21
HPO4-2 H+ ⇌ +PO4-3 KA=4.8X10-13 Pka=12.32
Therefore the most suitable of the three buffer system is HPO4-2/H2PO4– because the pKa of the acid is close to the desired pH
pH = pka + log(conjugate base)
(acid
7.40 =7.21 +log(HPO4-2)
(H2PO4–)
log(HPO4–2) = 0.19
H2PO4–
Taking the antilog we obtain
(HPO4-2) =10 0.19
(H2PO4–)
=1.5
Thus one way to prepare a buffer solution with a pH of 7.40 is to dissolve Disodium hydrogen phosphate(Na2HPO4) and Sodium dihydrogen phosphate (NaH2PO4) in a mole ratio of 1 .5:1.0 in water
18. A buffer solution was prepared which had a concentration of 0.20 mol dm–3in ethanoic acid and 0.10 mol dm–3 in sodium ethanoate. If the Ka for ethanoic acid is 1.74 x 10–5 mol dm–3, calculate the theoretical hydrogen ion concentration and pH of the buffer solution.
Ka = [H+(aq)] [salt(aq)]/[acid(aq)]
1.74 x 10–5 = [H+(aq)] x 0.10 / 0.20
[H+(aq)] = 1.74 x 10–5 x 0.20/0.10
= 3.48 x 10–5 mol dm–3
pH = –log(3.48 x 10–5)
= 4.46
20. In what ratio should a 0.30 mol dm–3of ethanoic acid be mixed with a 0.30 mol dm–3 solution of sodium ethanoate to give a buffer solution of pH 5.6?
Ka for ethanoic acid is 1.74 x 10–5 mol dm–3
[H+(aq)] = 10–pH = 10–5.6 = 2.51 x 10–6 mol dm–3
Ka = [H+(aq)] [salt(aq)]/[acid(aq)]
[salt]/[acid] = Ka/[H+(aq)] = 1.74 x 10–5/2.51 x 10–6
= 6.93 Therefore volume ratio is 6.93 : 1 for salt : acid, e.g. 6.93 cm3 of 0.30M sodium ethanoate is mixed with 1.0 cm3 of 0.30 M ethanoic acid to give a buffer solution of pH 5.6.
Note that the pH is determined by the ratio of concentrations, but the buffering capacity of the solution can be increased by increasing the concentrations of both components in the same molar concentration ratio.
21. What is the pH of a buffer solution made from dissolving 2.0g of benzoic acid and 5.0g of sodium benzoate in 250 cm3 of water?
Ka benzoic acid = 6.3 x 10–5 mol dm–3, Ar‘s: H = 1, C = 12, O = 16,
Na = 23
Molecular masses: Mr(C6H5COOH) = 122,
Mr(C6H5COO–Na+) = 144,
250 cm3 = 0.25dm3
moles acid C6H5COOH = 2.0/122 = 0.0164 mol, molarity = 0.0656 mol dm–3
moles salt C6H5COO–Na+ = 5.0/144 = 0.0347 mol, molarity = 0.139 mol dm–3
[H+(aq)] = Ka [acid(aq)]/[salt(aq)]
[H+(aq)] = 6.3 x 10–5 x 0.0656 / 0.139 = 2.97 x 10–5 mol dm–3
pH = –log[H+(aq)] = –log(2.97 x 10–5) = 4.53
Ka propanoic acid = 1.3 x 10–5 mol dm–3, total volume of buffer = 150 cm3
molarities in the mixture:
[salt] = 0.40 x 100/150 = 0.267 mol dm–3
[acid] = 0.20 x 50/150 = 0.0667 mol dm–3
[H+(aq)] = Ka [acid(aq)]/[salt(aq)]
[H+(aq)] = 1.3 x 10–5 x 0.0667 / 0.267
= 3.24 x 10–6 mol dm–3
pH = –log[H+(aq)]
= –log(3.24 x 10–6)
= 5.48
Using the Henderson equation
pHbuffer = pKa + log10([salt(aq)] / [acid(aq)])
22. Calculate the pH of buffer solution made by mixing together 100 cm3of 0.100M ethanoic acid and 50 cm3 of 0.400M sodium ethanoate,given that Ka for ethanoic acid is 1.74 x 10–5 mol dm–3
Now because the volumes are not equal, the real concentrations in the mixture must be worked out.
The total volume is 150 cm3, therefore the dilutions are given by
[acid] = 0.1 x 100/150 = 0.06667
[salt] = 0.4 x 50/150 = 0.1333
Substituting in the Henderson Equation gives
pHbuffer = –log10(1.74 x 10–5) + log10(0.1333/0.0667)
pHbuffer = –log10(1.74 x 10–5) + log10(2)
pHbuffer = 4.76 + 0.3010
pHbuffer = 5.06