Question

If 38.55 mL of HCl is required to titrate 2.150 g of Na2CO3 according to the following equation, what is the molarity of the HCl solution? Na2CO3 + 2HCl (aq)→ 2NaCl (aq) + CO2 (g) + H20 (1)

Answer 1

The molarity of the HCl solution : 1.052 M

Further explanation

Given

Volume of HCl = 38.55 ml

mass of Na₂CO₃ (MW=106 g/mol)=2.15 g

Reaction

Na₂CO₃(aq) + 2HCl (aq)→ 2NaCl (aq) + CO₂ (g) + H₂0 (l)

Required

The molarity of the HCl solution

Solution

mol Na₂CO₃ :

$\tt mol=\dfrac{mass}{MW}=\dfrac{2.15}{106}=0.02028$

From the equation, mol ratio Na₂CO₃ : HCl = 1 : 2, so mol HCl :

$\tt mol~HCl=2\times 0.02028=0.04056$

The molarity :

$\tt M=\dfrac{n}{V}=\dfrac{0.04056}{0.03855~L}=1.052~M$