Question

For a given mixture of $NaHC{O}_3$ and $N{a}_{2}C{O}_3$, volume of a given $HCl$ required is $x$ mL with phenolphthalein indicator and further $y$ mL required with methyl orange indicator. Hence, volume of $HCl$ for complete reaction of $NaHC{O}_3$ is :

• A. $2x$
• B. $\frac{x}{2}$
• C. $y$
• D. $(y-x)$

Answer 1

Answer: (D)

$(y-x)$

The phenolphthalein indicator shows the end point which corresponds to $50\%$ neutralization of sodium carbonate.
The remaining $50\%$ neutralization is given by methyl orange.
It also shows $100\%$ neutralization of sodium bicarbonate.
Let $V_1$ represent the volume of $HCl$ required for $100\%$ neutralization of sodium carbonate and $V_2$ represent the volume of $HCl$ required for $100\%$ neutralization of sodium bicarbonate.
Hence, volume of $HCl$ for complete reaction of sodium bicarbonate is $(y-x)=V_2+\frac{V_1}{2}-\frac{V_1}{2}=V_2$