Question

$1.22$ g sample of ${Na}_{2}C{O}_3$ and $K_{2}C{O}_3$ was dissolved in water to form $100$ mL of a solution. $20$ mL of this solution required $40$ mL of $0.1N$ $HCl$ for complete neutralization. Calculate the mass of ${Na}_{2}C{O}_3$ in the mixture. If another $20mL$ of this solution is treated with excess of $Ba{Cl}_2$, then the mass of pecipitate will be (write the nearest digit just after the decimal point as $0.456=5$) :

Answer 1

Let mass of ${Na}_{2}C{O}_3=a$ g and $K_{2}C{O}_3=b$ g.
$a+b=1.22...\left(i\right)$

For neutralization reaction of $100$ mL solution,
meq. of ${Na}_{2}C{O}_3+$ meq. of $K_{2}C{O}_3=$ meq. of $HCl$
$\frac{a}{\frac{106}{2}}\times 1000+\frac{b}{\frac{138}{2}}\times 1000=\frac{40\times 0.1\times 100}{20}$
$63a+53b=73.14...\left(ii\right)$

By Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$a=0.53$ g
$b=0.69$ g

Further solution of ${Na}_{2}C{O}_3+K_{2}C{O}_3$ gives ppt. of $BaC{O}_3$ with $Ba{Cl}_2$.
Meq. of $BaC{O}_3$ $=$ Meq. of ${Na}_{2}C{O}_3+$ Meq. of $K_{2}C{O}_3$ (in $20$ mL)
$=$ Meq. of $HCl$ for $20$ mL mixture = $40\times 0.1=4$
Therefore, $\frac{w}{\frac{197}{2}}\times 1000=4$

Mass of $BaC{O}_3=0.394$ g

So, nearest digit after decimal point is $4$.