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

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.

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