Textbook Problem

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What volume of 0.955 M HCl, in milliliters, is required to titrate 2.152 g of Na₂CO₃ to the equivalence point?

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

Interpretation Introduction

Interpretation:

The volume of $0.955MHCl$ required to titrate $2.152g$ of $N{a}_{2}C{O}_3$ to the equivalence point has to be calculated in milliliters.

Concept introduction:

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• $\text{Concentration}\text{of}\text{substance}\text{=}\frac{\text{Amount}\text{of substance}}{\text{volume}\text{of}\text{the}\text{substance}}$
• $\text{Amount}\text{of substance}=\text{Concentration}\text{of}\text{substance}\times \text{volume}\text{of}\text{the}\text{substance}$
• $\text{Volume}\text{of}\text{the}\text{substance}=\frac{\text{Amount}\text{of substance}}{\text{Concentration}\text{of}\text{substance}}$

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $N{a}_{2}C{O}_3{}_{(aq)}+2HC{l}_{(aq)}\to H_2{O}_{(l)}+C{O}_2{}_{(g)}+2NaC{l}_{(aq)}$

The amount (moles) of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ available can be calculated as follows,

  $2.152g{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{1mol{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{106g}=0.02030mol{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

The balanced equation for the reaction shows that $\text{1}\text{mol}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ requires $2\text{mol}$ of $HCl$.  This is the required stoichiometric factor to obtain the amount of $HCl$ present.

Thus the amount of $HCl$ can be calculated as follows,

  $\text{0}$