Question

1 L solution of $N{a}_{2}C{O}_3$ and $NaOH$ was made in $H_{2}O$. $100$ mL of this solution required $20$ mL of $0.4M$ $HCl$ in the presence of phenolphthalein. However, another $100$ mL sample of the same solution required $25$ mL of the same acid in the presence of methyl orange as an indicator.

If molar ratio of $N{a}_{2}C{O}_3$ and $NaOH$ in the original mixture is $\frac{x}{y}$, then $x+y$ is :

Answer 1

Let $a$ and $b$ meq. of $N{a}_{2}C{O}_3$ and $NaOH$ are present in the mixture respectively.
When phenolphthalein is used as indicator,
$\frac{a}{2}$ meq. of $N{a}_{2}C{O}_3+b$ meq. of $NaOH=$ meq. of $HCl$

$\frac{a}{2}+b=20\times 0.4\times 1=8$ .......(i)

When methyl orange is used as indicator,

$a$ meq. of $N{a}_{2}C{O}_3+b$ meq. of $NaOH=$ meq. of $HCl$
$a+b=25\times 0.4\times 1=10$ ....(ii)
Solving for $a$ and $b$ from equation (i) and (ii), we get,
$a=4$ meq. $=\frac{4}{2}$mmol $=2$ mmol

$b=6$ meq. $=\frac{6}{1}$ mmol $=6$ mmol

Ratio of $N{a}_{2}C{O}_3:NaOH=2:6=1:3$
So, the value of $x+y$ is $3+1=4$.