Solved problem

Sodium carbonate (Na₂CO₃) is used to determine the concentration of aqueous HCl. 21.2

of aqueous HCl reacts with 0.130 g Na₂CO₃ as shown. What is the concentration of HCl in the solution?

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O + CO₂(aq)

m(Na₂CO₃)
Known

n	=	m
		M

→
Step 1

n(Na₂CO₃)

n(Na₂CO₃)	=	n(HCl)
1	=	2

→
Step 2

n(HCl)

c	=	n
		V

→
Step 3

c(HCl)
Unknown

Step 1: Use given information (mass of a pure substance) to calculate n(known)

n(Na₂CO₃)	=	0.130 g
		106.0 g mol^–1

n(Na₂CO₃) = 1.22642 × 10^–3 mol n(HCl) = 2.45283 × 10^–3 mol

Step 2: Calculate n(unknown) from n(known) and the coefficients on these substances in the balanced equation for the reaction.

1.22642 × 10^–3 mol	=	n(HCl)
1	=	2

n(HCl) = 2.45283 × 10^–3 mol

Step 3: Use calculated n(unknown) and V(unknown) to calculate c(unknown).

c(HCl)	=	2.452 × 10^–3 mol
		21.21 × 10^–3

c(HCl) = 0.1160