Chemical Formulae, Equations and Calculations Part 2
5 marks · EasyA student investigates the reaction between barium hydroxide and sulfuric acid:
Calculate the number of moles of barium hydroxide in 40.0 cm³ of 0.200 mol/dm³ barium hydroxide solution.
During a titration, 25.0 cm³ of the barium hydroxide solution from part (a) is neutralised by 19.5 cm³ of sulfuric acid solution.
Calculate the concentration of the sulfuric acid in mol/dm³. Show your working.
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) 0.00800 mol · must show Conversion of 40.0 cm³ to 0.0400 dm³, Application of: moles = volume (dm³) × concentration
(b) 0.205 mol/dm³ · must show Find moles in 25.0 cm³ Ba(OH)₂, ratio to H₂SO₄, Divide moles by 0.0195 dm³ to get concentration
Common mistakes: not converting cm³ to dm³; incorrect use of stoichiometric ratio; using wrong volume in final division
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Converts 40.0 cm³ to 0.0400 dm³ OR states moles = volume × concentration
A1 (a) · 1 mark · Correctly calculates moles of Ba(OH)₂: 0.00800 mol.
M2 (b) · 1 mark · Calculates moles in 25.0 cm³ of Ba(OH)₂; sets up stoichiometric relationship (recognises 1:1 ratio with H₂SO₄). (follow-through)
M3 (b) · 1 mark · Divides moles of H₂SO₄ by correct volume in dm³ to find concentration. (follow-through)
A2 (b) · 1 mark · Correct final answer for concentration: 0.205 mol/dm³ (allow rounding). (follow-through)
Worked solution
a) Convert volume from cm³ to dm³:
Calculate moles of Ba(OH)₂:
b) Titration: 25.0 cm³ of Ba(OH)₂ neutralised by 19.5 cm³ of H₂SO₄.
Moles of Ba(OH)₂ in 25.0 cm³:
Balanced equation shows 1 mol Ba(OH)₂ reacts with 1 mol H₂SO₄.
So, moles of H₂SO₄ used = 0.00500 mol.
Volume of H₂SO₄ used = 19.5 cm³ = 0.0195 dm³.
Concentration of H₂SO₄ = moles / volume:
Answer to 3 significant figures: 0.205 mol/dm³.