Chemistry Volume and Concentration question

Question

a) 100cm³ of a standard solution of Na₂CO₃ was produced by dissolving 0.537g of Na₂CO₃ in water in a volumetric flask. Calculate the concentration, in mol dm^-3, of the Na₂CO₃ solution.

b) On titration, 25.00cm³ of this Na₂CO_{3 (aq)} solution required 23.75cm³ of the hydrochloric acid for complete neutralisation. Calculate the concentration, in mol dm^-3, of the hydrochloric acid.

c) Calculate the mass, in grams, of HCl in 560cm³ of hydrochloric acid of concentration 0.145mol dm^-3.

Answer 1

Answers first: (a) 0.0507 mol dm-3, (b) 0.107 mol dm-3, (c) 2.96 g. Full working below, and the two traps this question is built around are flagged as they come up.

Every part runs off the same three relationships. Write them at the top of your page before you touch the numbers:

n = c x V          (V must be in dm3, not cm3)
n = mass / molar mass
1 dm3 = 1 L = 1000 cm3

"mol dm-3" is just the SI way of writing molarity, so a 0.100 mol dm-3 solution and a 0.100 M solution are the same thing. Dividing every cm3 figure by 1000 as your first move will save you more marks on this topic than anything else.

(a) Concentration of the standard Na2CO3 solution

Molar mass of Na2CO3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol.

n(Na2CO3) = 0.537 g / 105.99 g/mol = 5.067 x 10^-3 mol
V         = 100 cm3 = 0.100 dm3
c         = 5.067 x 10^-3 / 0.100 = 0.0507 mol dm-3

(b) Concentration of the hydrochloric acid

You need the balanced equation before any titration arithmetic:

Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2

That 2 is the single commonest lost mark on this question. Carbonate soaks up TWO protons, so the acid is consumed at twice the rate of the carbonate. Forget it and you get 0.0533, which is exactly half the right answer, and half-answers are hard to spot as wrong.

n(Na2CO3) in 25.00 cm3 = 0.0507 x 0.02500 = 1.267 x 10^-3 mol
n(HCl)                 = 2 x 1.267 x 10^-3 = 2.533 x 10^-3 mol
c(HCl)                 = 2.533 x 10^-3 / 0.02375 = 0.107 mol dm-3

(c) Mass of HCl in 560 cm3 of 0.145 mol dm-3 acid

Trap number two: 0.145 is not the answer you just calculated in (b). Part (c) is a fresh, unconnected problem that happens to sit under the same question number. Carrying 0.107 forward here is a very common slip. Read each part as if it arrived on its own.

n(HCl) = 0.145 mol dm-3 x 0.560 dm3 = 0.0812 mol
M(HCl) = 1.01 + 35.45 = 36.46 g/mol
mass   = 0.0812 x 36.46 = 2.96 g

Sanity check on (b), which is worth doing every time. The carbonate is about 0.05 mol dm-3 and the acid came out about 0.107, so the acid is roughly twice as concentrated. It also reacts in a 2:1 ratio. Those two factors cancel, which is exactly why the titre (23.75 cm3) came out so close to the aliquot (25.00 cm3). If your answer had come out near 0.05 or near 0.2, the arithmetic would be telling you the volumes should have been very different, and they were not.

Significant figures. The volumes are given to 4 sig figs but the mass 0.537 g is only 3, so 3 sig figs is the ceiling on every answer. Do not round in the middle; carry the extra digits and round once at the end.

One practical point behind the phrase "complete neutralisation." A carbonate titration goes in two steps: CO3 2- to HCO3- first, then HCO3- to H2CO3 (which leaves as CO2 and water). Phenolphthalein changes colour at the END of the FIRST step and would give you a titre for a 1:1 ratio, not 2:1. To reach complete neutralisation you use methyl orange, and the reliable version of this experiment boils the solution near the endpoint to drive off dissolved CO2, which otherwise makes the colour change drift. That is why the standard method specifies the indicator rather than leaving it to you.