Sodium bicarbonate (baking soda), NaHCO₃, can be purified by dissolving it in hot water (60°C), filtering to remove insoluble impurities, cooling to 0 °C to precipitate solid NaHCO₃ , and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO₃ that remains in solution is not recovered. The solubility of NaHCO₃ in hot water of 60 °C is 164 g/L. Its solubility in cold water of 0 °C is 69 g/L. What is the percent yield of NaHCO₃ when it is purified by this method?

It is given that the solubility of sodium bicarbonate in hot water is 164 g/L which is the theoretical yield of \({\rm{NaHC}}{{\rm{O}}_3}\)

Now, we know that solubility means the solute is dissolved in the solution, and in cold water the \({\rm{NaHC}}{{\rm{O}}_3}\) is present along with the other impurities, so the amount of \({\rm{NaHC}}{{\rm{O}}_3}\) that is present in the solution or the actual yield of \({\rm{NaHC}}{{\rm{O}}_3}\) will be 164 – 69 = 95 g/L

The net percentage yield of the compound is calculated by:

\({\rm{Percentage yield = }}\frac{{{\rm{Actual yield}}}}{{{\rm{Theoretical yield}}}}{\rm{ }} \times {\rm{ 100}}\)

Therefore, the percentage yield of the \({\rm{NaHC}}{{\rm{O}}_3}\) is

\({\rm{Percentage yield = }}\frac{{95}}{{164}}{\rm{ }} \times {\rm{ 100 = 58 percent}}\)