The standard calculation
4 marks · NormalEthanoic acid ( ) reacts with sodium hydroxide ( ) according to the equation:
A student titrates 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide with ethanoic acid. On average, 21.0 cm³ of ethanoic acid is needed to neutralise the alkali.
State the ratio in which sodium hydroxide reacts with ethanoic acid.
Calculate the number of moles of sodium hydroxide present in 25.0 cm³.
Calculate the number of moles of ethanoic acid used in the reaction.
Calculate the concentration of the ethanoic acid.
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) 1:1
(b) 0.00500 mol
(c) 0.00500 mol
(d) 0.238 mol/dm³
Common mistakes: incorrectly identifying the ratio; not converting cm³ to dm³
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States 1:1 ratio for : from the equation.
M1 (b) · 1 mark · Correctly calculates moles of : mol.
M2 (c) · 1 mark · Uses the mole ratio (1:1) to deduce moles of ethanoic acid used = 0.00500 mol. (follow-through)
A1 (d) · 1 mark · Correct calculation of concentration: answer = 0.238 mol/dm³. (follow-through)
Worked solution
(a) From the equation, the mole ratio is 1:1.
(b) Convert 25.0 cm³ to dm³: 25.0 ÷ 1000 = 0.0250 dm³ Moles of NaOH = 0.200 × 0.0250 = 0.00500 mol
(c) As ratio is 1:1, moles of ethanoic acid used = 0.00500 mol
(d) Volume of ethanoic acid = 21.0 cm³ = 0.0210 dm³ Concentration = 0.00500 ÷ 0.0210 = 0.238 mol/dm³