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Molarity
standard solutions are commonly expressed in terms of molar concentrations or molarity (M). Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution
Therefore ,standard solutions are specified in terms of the number of moles of solute dissolved in 1 litre of solution; for any solution,
- Molarity (M) = Moles of solute
- Volume of solution in litres
Example
- If 5.00g of sodium chloride is dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,
- (a) what is the concentration in g/dm3?
- Volume = 250/1000 = 0.25 dm3
- Concentration = mass / volume
- = 5/0.25
- = 20 g/dm3
What is the molarity of the solution?
- Ar(Na) = 23, Ar(Cl) = 35.5,
- so Mr(NaCl) = 23 + 35.5 = 58.5
- mole NaCl = 5.0/58.5 = 0.08547
- volume = 250/1000 = 0.25 dm3
- molarity = mol of solute / volume of solvent
- Molarity = 0.08547/0.25
- = 0.342 mol/dm3
5.95g of potassium bromide was dissolved in 400cm3 of water.
- (a) Calculate its molarity. [Ar‘s: K = 39, Br = 80]
- moles = mass / formula mass, (KBr = 39 + 80 = 119)
- mol KBr = 5.95/119 = 0.050 mol
- 400 cm3 = 400/1000 = 0.400 dm3
- molarity = moles of solute / volume of solution
- molarity of KBr solution = 0.050/0.400 = 0.125 mol/dm3
- (b) What is the concentration in grams per dm3?
- concentration = mass / volume, the volume = 400 / 1000 = 0.4 dm3
- concentration = 5.95 / 0.4 = 14.9 g/dm3
What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3) of a 0.500 mol dm-3 (0.5M) solution? [Ar‘s: Na = 23, O = 16, H = 1]
- 1 mole of NaOH = 23 + 16 + 1 = 40g
- molarity = moles / volume, so mol needed = molarity x volume in dm3
- 500 cm3 = 500/1000 = 0.50 dm3
- mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH
- therefore mass = mol x formula mass
- = 0.25 x 40 = 10g NaOH required
- How many moles of H2SO4 are there in 250 cm3 of a 0.800 mol dm-3 (0.8M) sulphuric acid solution?
What mass of acid is in this solution?
- [Ar‘s: H = 1, S = 32, O = 16]
- (a) molarity = moles / volume in dm3, rearranging equation for the sulfuric acid
- mol H2SO4 = molarity H2SO4 x volume of H2SO4 in dm3
- mol H2SO4 = 0.800 x 250/1000 = 0.200 mol H2SO4
- (b) mass = moles x formula mass
- formula mass of H2SO4 = 2 + 32 + (4×16) = 98
- 0.2 mol H2SO4 x 98 = 19.6g of H2SO4
A solution of calcium sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
- Calculate the concentration in (a) g/dm3, (b) g/cm3 and (c) mol/dm3.
- (a) concentration = 0.500/2.00 = 0.250 g/dm3,
- then since 1dm3 = 1000 cm3
- (b) concentration = 0.250/1000 = 0.00025 g/cm3
- (or from 0.500/2000)
- (c) At. masses: Ca = 40, S = 32, O = 64,
- formula mass CaSO4 = 40 + 32 + (4 x 16) = 136
- moles CaSO4 = 0.5 / 136 = 0.00368 mol in 2.00 dm3 of water
- concentration CaSO4 = 0.00368 / 2
- = 0.00184 mol/dm3
Error in Titration Calculations
Different methods are used to determine the equivalence point of a titration. No matter which method is used, some error is introduced, so the concentration value is close to the true value, but not exact. For example, if a colored pH indicator is used, it might be difficult to detect the color change. Usually, the error here is to go past the equivalence point, giving a concentration value that is too high.
Another potential source of error when an acid-base indicator is used is if water used to prepare the solutions contains ions that would change the pH of the solution. For example, if hard tap water is used, the starting solution would be more alkaline than if distilled deionized water had been the solvent.
If a graph or titration curve is used to find the endpoint, the equivalence point is a curve rather than a sharp point. The endpoint is a sort of “best guess” based on the experimental data.
The error can be minimized by using a calibrated pH meter to find the endpoint of an acid-base titration rather than a color change or extrapolation from a graph.