Question

A

1 g

sample containing

NaOH

as the only basic substance and some inert impurity was left exposed to atmosphere for a very long time so that part of

NaOH

got converted into

{Na}_{2} {CO}_{3}

by absorbing

{CO}_{2}

from atmosphere. The resulting sample was dissolved in water and volume made upto

100 mL

. A

20 mL

portion of this solution required

16 mL 0.25 M HCl

solution to reach the equivalence point when methyl orange was used as indicator. In a separate analysis,

20 mL

portion of the same solution was taken along with phenolphthalein indicator and mixed with

50 mL

of

0.1 M HCl

solution. An additional

9.00 mL 0.1 M

Ba (OH)_{2}

solution was required to just restore the pink colour of solution. Determine mass percentage of

NaOH

in the original sample and mass percentage of

{Na}_{2} {CO}_{3}

in the sample after exposure to atmosphere.

Accepted Answer

In presence of methyl orange, the whole $NaOH$ and ${Na}_{2} {CO}_{3}$ are neutralized

$\Rightarrow$ meq of $HCl = 16 0.25 = 4 =$ meq of $NaOH$ original

$\Rightarrow$ Total meq of $NaOH$ in original $1.0 g$ sample

$20 mL$

$\Rightarrow$ mass $%$ of $NaOH$ (original) $= \frac{20 \times 40 \times 100}{1000} = 80$
Now, let us assume that in $20 mL, xm mol$ of $NaOH$ has got converted to ${Na}_{2} {CO}_{3}$

$\Rightarrow$ In $20 mL, m$ mol of $NaOH = 4 x$

$m mol$ of ${Na}_{2} {CO}_{3} = \frac{x}{2}$
In $2$ nd titration, $NaOH$ used in titration of $NaOH$

$\Rightarrow$ upto phenolphthalein end point, $m$ mol of $HCl$

required $20 mL$

$\Rightarrow x = 1.6$

$\Rightarrow$ Total $Ba (OH)_{2}$ formed $= \frac{x}{2} 5$

$= \frac{5 x}{2} = 4$

$m mol$ of $NaOH$ left unreacted $= (20 + 4) \div 2 = 12$

$\Rightarrow$ weight of $1.0 g$ of exposed sample

$= 1 - \frac{8 \times 40}{1000} + \frac{4 \times (106 + 18)}{1000} = 1.176 g$

$\Rightarrow$ weight $%$ of ${Na}_{2} {CO}_{3}$ in exposed sample

$= \frac{4 \times 106}{1000 \times 1.176} \times 100 = 36.05 %$

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