Changing the concentration of the reactants

5 marks · Hard

A student prepares dilute hydrochloric acid for a reaction by adding water to a known volume of 2.50 mol/dm³ hydrochloric acid. She then uses the solution to react with marble chips.

(a)

She mixes 10.0 cm³ of 2.50 mol/dm³ acid with 30.0 cm³ of water to make 40.0 cm³ of solution. Calculate the concentration, in mol/dm³, of the hydrochloric acid after dilution.

[2 marks]

(b)

She repeats the process, using 25.0 cm³ of 2.50 mol/dm³ acid topped up with water to 40.0 cm³. Calculate the new concentration, in mol/dm³, after dilution.

[2 marks]

(c)

State the formula used to calculate the final concentration when diluting a solution.

[1 mark]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

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(a) 0.625 mol/dm³

(b) 1.56 mol/dm³

(c) final concentration = (volume of acid / total volume) × original concentration · must show mentions (volume of acid / total volume) × original concentration

Also accept: Alternative layout of the dilution equation accepted.

Common mistakes: Incorrectly using only the water volume instead of total volume.; Forgetting to express answer in mol/dm³.

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · States or uses correct formula: (volume of acid / total volume) × original concentration, or equivalent.

A1 (a) · 1 mark · Correct final answer for part (a): 0.625 mol/dm³ (allow 2 or 3 sig. figs and minor rounding error).

M1 (b) · 1 mark · Correct application of the dilution formula to part (b). (follow-through)

A1 (b) · 1 mark · Correct final answer for part (b): 1.56 mol/dm³ (allow 2 or 3 sig. figs and minor rounding error). (follow-through)

B1 (c) · 1 mark · States the correct dilution formula in words or symbols: (volume of acid / total volume) × original concentration.

Worked solution

(a) Use the formula: concentration after dilution = (volume of acid / total volume) × original concentration Substitute: (10.0 / 40.0) × 2.50 mol/dm³ = 0.25 × 2.50 = 0.625 mol/dm³

(b) (25.0 / 40.0) × 2.50 mol/dm³ = 0.625 × 2.50 = 1.5625 ≈ 1.56 mol/dm³