The standard calculation
4 marks · NormalA sample of nitric acid ( ) reacts with sodium carbonate ( ) solution. The balanced equation is:
A student titrates 25.0 cm³ of 0.150 mol/dm³ sodium carbonate. On average, 36.0 cm³ of nitric acid is required to completely react.
According to the equation, what is the ratio of moles of nitric acid to moles of sodium carbonate required?
Calculate the number of moles of sodium carbonate used.
Determine the number of moles of nitric acid required to react completely.
Calculate the concentration of nitric 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) 2:1
(b) 0.00375 mol
(c) 0.00750 mol
(d) 0.208 mol/dm³
Common mistakes: wrong ratio taken from equation; wrong conversion of units
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States correct ratio: 2 mol : 1 mol (2:1).
M1 (b) · 1 mark · Calculates moles: 0.150 × 0.0250 = 0.00375 mol .
M2 (c) · 1 mark · Uses 2:1 ratio to get 0.00375 × 2 = 0.00750 mol . (follow-through)
A1 (d) · 1 mark · Finds concentration of : 0.0075 ÷ 0.0360 = 0.208 mol/dm³. (follow-through)
Worked solution
(a) From the equation, 2 moles of nitric acid react with 1 mole of sodium carbonate. Ratio = 2:1.
(b) 25.0 cm³ = 0.0250 dm³; moles = 0.150 × 0.0250 = 0.00375 mol
(c) Moles = 0.00375 × 2 = 0.00750 mol
(d) 36.0 cm³ = 0.0360 dm³; concentration = 0.00750 ÷ 0.0360 = 0.208 mol/dm³