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# 1.20g sample of $Na_2CO_3$ and $K_2CO_3$ was dissolved in water to form 100ml of a solution.20ml of this solution required 40ml of 0.1N HCl for complete neutralization.Calculate the weight of $Na_2CO_3$ in mixture. If another 20 ml of this solution is treated with excess of $BaCl_2$,what will be the weight of precipitate?

$\begin{array}{1 1}(a)\;0.5962g,0.394g&(b)\;0.234g,0.142g\\(c)\;0.1243g,0.235g&(d)\;0.42g,0.24g\end{array}$

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A)
$Na_2CO_3=ag$
$K_2CO_3=bg$
$a+b=1.20$------(1)
For neutralization reaction of 100ml solution.
Meq of $Na_2CO_3$+Meq of $K_2CO_3$=Meq of HCl
$\large\frac{a}{\Large\frac{106}{2}}$$\times 1000+\large\frac{b}{\Large\frac{138}{2}}$$\times 1000=\large\frac{40\times 0.1\times 100}{20}$
$69a+53b=73.14$-------(2)
From (1) & (2)
$a=0.5962g$
$b=0.6038g$
$Na_2CO_3+K_2CO_3$ gives ppt of $BaCO_3$ with $BaCl_2$
Meq of $BaCO_3$=Meq of $Na_2CO_3$+Meq of $K_2CO_3$
$\Rightarrow$ Meq of HCl for 20ml mixture.
$\Rightarrow 40\times 0.1=4$
$\therefore \large\frac{w}{\Large\frac{197}{2}}$$\times =4$
$\therefore$ wt of $BaCO_3=0.394g$
Hence (a) is the correct option.