A solution contains $ ext{Na}_2 ext{CO}_3$ and $ ext{NaHCO}_3$. $10 ext{ mL}$ of the solution r

Question

A solution contains $	ext{Na}_2	ext{CO}_3$ and $	ext{NaHCO}_3$. $10 	ext{ mL}$ of the solution requires $2.5 	ext{ mL}$ of $0.1 	ext{ M H}_2	ext{SO}_4$ for neutralization using phenolphthalein as an indicator. Methyl orange is then added, and a further $2.5 	ext{ mL}$ of $0.2 	ext{ M H}_2	ext{SO}_4$ is required to reach the end point. Which of the following statements is/are correct about the components in ONE litre of the original solution?

Professor PC Thomas Classes · Answer

Phenolphthalein endpoint (half-neutralization of carbonate): $2	ext{Na}_2	ext{CO}_3 + 	ext{H}_2	ext{SO}_4 ightarrow 2	ext{NaHCO}_3 + 	ext{Na}_2	ext{SO}_4$. 
Moles of $	ext{H}_2	ext{SO}_4 = 2.5 	imes 0.1 = 0.25$ mmol. Moles of $	ext{Na}_2	ext{CO}_3$ in 10 mL = $0.25 	imes 2 = 0.5$ mmol. 
Methyl orange endpoint (neutralizes all bicarbonate, including what was formed): $2	ext{NaHCO}_3 + 	ext{H}_2	ext{SO}_4 ightarrow 	ext{Na}_2	ext{SO}_4 + 2	ext{CO}_2 + 2	ext{H}_2	ext{O}$. 
Moles of $	ext{H}_2	ext{SO}_4 = 2.5 	imes 0.2 = 0.5$ mmol. Total $	ext{NaHCO}_3$ neutralized = $0.5 	imes 2 = 1.0$ mmol. 
Initial $	ext{NaHCO}_3$ + formed $	ext{NaHCO}_3 = 1.0 \implies b + 0.5 = 1.0 \implies b = 0.5$ mmol. 
In 1L (100 times 10 mL): Moles of both = $50$ mmol = $0.05$ mol (Molarity is $0.05 	ext{ M}$). 
Mass $	ext{Na}_2	ext{CO}_3 = 0.05 	imes 106 = 5.3$ g. Mass $	ext{NaHCO}_3 = 0.05 	imes 84 = 4.2$ g.