The standard calculation
3 marks · EasyA student uses 28.0 cm³ of hydrochloric acid to neutralise 25.0 cm³ of sodium hydroxide solution with a concentration of 0.120 mol/dm³. The equation for the reaction is:
Calculate the moles of sodium hydroxide used.
State how many moles of hydrochloric acid react with this amount, using the equation.
Calculate the concentration of hydrochloric acid in mol/dm³.
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.
Show answer
(a) 0.00300 mol
(b) 0.00300 mol
(c) 0.107 mol/dm³
Common mistakes: Incorrectly converting volume to dm³; Wrong stoichiometry
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Calculates moles of NaOH: (25.0/1000) × 0.120 = 0.00300 mol.
M2 (b) · 1 mark · States moles HCl equals moles NaOH from 1:1 ratio in equation (0.00300 mol). (follow-through)
A1 (c) · 1 mark · Calculates concentration HCl: 0.00300/(28.0/1000) = 0.107 mol/dm³. (follow-through)
Worked solution
a) Moles NaOH = (25.0/1000) × 0.120 = 0.00300 mol
b) Reaction is 1:1 so moles HCl = 0.00300 mol
c) Concentration HCl = moles/volume in dm³ = 0.00300 / (28.0/1000) = 0.107 mol/dm³