0-2 gm of a solution of mixture of NaOH and Na2CO2and inert impurities was first titrated with phenolphthalein and N-10 HCl 17-5 ml of HCl was required at the end point- nbsp- After this methyl orange was added and 2-5 ml of same HCl was again required for next end point- nbsp- Find out percentage of NaOH and Na2CO3 in the mixture-

Dear Student, In the first titration when we use phenolphthalein , nbsp;NaOH will be neutralized by HCl nbsp; so nbsp; (NV)HCl= (NV)NaOH so nbsp; n ...

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Standard XII

Chemistry

Lewis Theory

0.2 gm of a s...

Question

0.2 gm of a solution of mixture of NaOH and Na2CO2and inert impurities was first titrated with phenolphthalein and N/10 HCl 17.5 ml of HCl was required at the end point. After this methyl orange was added and 2.5 ml of same HCl was again required for next end point. Find out percentage of NaOH and Na2CO3 in the mixture.

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Solution

Dear Student,
In the first titration when we use phenolphthalein , NaOH will be neutralized by HCl
so
(NV)_HCl= (NV)_NaOH
so milliequivalents of NaOH=(NV)_NaOH = $0.1 \times 17.5$ =1.75
equivalents of NaOH = $\frac{1.75}{1000}$
weight of NaOH = equivalents $\times e q u i v a l e n t w e i g h t$
weight of NaOH = 0.7 gm
% NaOH = $\frac{0.07}{0.2} \times 100$ = 35%
similarly in the second titration we use methyl orange
(NV)_HCl=(NV)_Na2CO3
(NV)Na2CO3= $2.5 \times 0.1$
so milliequivalents of Na₂CO₃ =0.25
equivalents of Na₂CO₃ = $\frac{0.25}{1000}$
weight of Na₂CO₃ in the sample = equivalents $\times$ equivalent weight
(equivalent weight of Na₂CO₃ = $\frac{0.25}{1000} \times 106$
weight of Na₂CO₃in the mixture = 0.0265 gm
% of Na₂CO₃= $\frac{0.0265}{0.2} \times 100$ = 13.25%
%NaOH =35%
% $N a_{2} C O_{3}$ = 13.5%
Regards!

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