Chapter 11, Problem 62A

Interpretation Introduction

Interpretation: If one tablet containing 0.0119 moles of sodium hydrogen carbonate is dissolved, the moles of sodium citrate forms needs to be determined.

  ${\text{3 NaHCO}}_{\text{3(aq)}}{\text{ + H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ 3 CO}}_{\text{2(g)}}{\text{ + 3 H}}_{\text{2}}{\text{O}}_{\text{(l) }}{\text{ + Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$

  ${\text{3 NaHCO}}_{\text{3(aq)}}{\text{ + H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ 3 CO}}_{\text{2(g)}}{\text{ + 3 H}}_{\text{2}}{\text{O}}_{\text{(l) }}{\text{ + Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$

Concept Introduction: The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of the mole concept can be written as:

  $\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 62A

  $\text{3}\text{.97}\times {\text{10}}^{\text{-3}}{\text{ moles Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$

Explanation of Solution

The reaction of sodium hydrogen carbonate with citric acid to form sodium citrate with carbon dioxide and water is as given below:

  ${\text{3 NaHCO}}_{\text{3(aq)}}{\text{ + H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ 3 CO}}_{\text{2(g)}}{\text{ + 3 H}}_{\text{2}}{\text{O}}_{\text{(l) }}{\text{ + Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$

The mole ratio relates the moles of the reactant with other reactant and product molecules in a balanced chemical equation.

Moles of sodium hydrogen carbonate = 0.0119 moles

3 moles of sodium hydrogen carbonate = 1 mole of sodium citrate

Calculate moles of sodium citrate with the help of mole ratio:

  $\text{0}{\text{.0119 molesNaHCO}}_{\text{3(aq)}}\text{ ×}\frac{{\text{1 mole Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}}{{\text{3 moles NaHCO}}_{\text{3(aq)}}}\text{= 3}{\text{.97×10}}^{\text{-3}}{\text{ moles Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$

Conclusion

Thus, $\text{3}\text{.97}\times {\text{10}}^{\text{-3}}{\text{ moles Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7(aq)}}$ are formed.