A mixture of NaOH and $\mathrm{Na}_2 \mathrm{CO}_3$ required 25 mL of 0.1 M HCl using phenolphthalei

Question

A mixture of NaOH and $\mathrm{Na}_2 \mathrm{CO}_3$ required 25 mL of 0.1 M HCl using phenolphthalein as the indicator. However, the same amount of the mixture required 30 mL of 0.1 M HCl when methyl orange was used as the indicator. The molar ratio of NaOH and $\mathrm{Na}_2 \mathrm{CO}_3$ in the mixture was:

Professor PC Thomas Classes · Answer

(c) Phenolphthalein gives the end point corresponding to the reactions : $$ \begin{aligned} \mathrm{NaOH}+\mathrm{HCl} & \longrightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \ \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{HCl} & \longrightarrow \mathrm{NaHCO}_3+\mathrm{NaCl} \end{aligned} $$ ∴ m-moles of NaOH + m-moles of $\mathrm{Na}_2 \mathrm{CO}_3$ $$ =\mathrm{m} 	ext {-moles of } \mathrm{HCl}=2.5 $$ Methyl orange gives the end point corresponding to the reactions : $$ \begin{aligned} \mathrm{NaOH}+\mathrm{HCl} & \longrightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \ \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \longrightarrow & 2 \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \ 	herefore 	ext { m-moles of } \mathrm{NaOH} & \end{aligned} $$ m-moles of $\mathrm{Na}_2 \mathrm{CO}_3=0.5$ m -moles of $\mathrm{NaOH}=2.5-0.5=2$ Ratio of m-moles of NaOH and $\mathrm{Na}_2 \mathrm{CO}_3$ $$ =\frac{2}{0.5}=4: 1 $$