# A mixture of KOH and $Na_2CO_3$ solution required 15ml of N/20 HCl using phenolphthalein as indicator .The same amount of alkali mixture when titrated using methyl orange as an indicator required 25ml of same acid.Calculate the amount of KOH and $Na_2CO_3$ present in solution.

$\begin{array}{1 1}(a)\;0.02g\;KOH,0.06g\;Na_2CO_3&(b)\;0.018g\;KOH,0.03g\;Na_2CO_3\\(c)\;0.014g\;KOH,0.053g\;Na_2CO_3&(d)\;0.018g\;KOH,0.06g\;Na_2CO_3\end{array}$

Meq of HCl used for mixture using phenolphthalein=$15\times \large\frac{1}{20}$
Meq of KOH+$\large\frac{1}{2}$ meq of $Na_2CO_3=\large\frac{3}{4}$-------(1)
Meq of HCl used for mixture using methyl orange=$25\times \large\frac{1}{20}$
(i.e) Meq of KOH+meq of $NA_2CO_3=\large\frac{5}{4}$------(2)
By equation (1) & (2)
$\large\frac{1}{2}$ Meq of $Na_2CO_3=(\large\frac{5}{4}-\frac{3}{4})$
Meq of $Na_2CO_3=1$
$\large\frac{w}{\Large\frac{106}{2}}$$\times 1000=1 w=0.053gNa_2CO_3 Meq of KOH=\large\frac{5}{4}$$-1=\large\frac{1}{4}$
$\large\frac{w}{\Large\frac{56}{1}}$$\times 1000=\frac{1}{4}$
$w=0.014gKOH$
Hence (c) is the correct answer.