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Hydrochloric Acid

A solution co...

Question

A solution contains ${N a}_{2} C O_{3}$ and $N a O H$ . Using phenolphthalein as an indicator, $25$ mL of a mixture requires $19.5$ mL of $0.005 N H C l$ for the end point. If methyl orange is the indicator, then $25$ mL of the solution requires $25.9$ of the same $H C l$ for end point. The concentration of $N a O H$ substance in g per litre is (write it as $0.23 = 2$ (only single digit)) :

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Solution

Meq. of $H C l$ for $35$ mL mixture using phenolphthalein indicator = $19.5 \times 0.005 = 0.0975$
$∴$ Meq. of $N a O H +$ $\frac{1}{2}$ Meq. of ${N a}_{2} C O_{3} = 0.0975$ ...(i)
Now, meq. of $H C l$ used for $25$ mL mixture using methyl orange as indicator $= 25.9 \times 0.005 = 0.1295 . . . (i i)$
$∴$ By eqs. $(i)$ and $(i i)$ , we get meq. of $N a O H$ $= 0.0655$
$\frac{w}{40} \times 1000 = 0.0655$
$∴$ Mass of $N a O H =$ $2.62 \times 10^{- 3} g / 25$
Strength of NaOH = $0.1048 g / l i t r e$
and answer is $1$ .
Also,
Meq. of ${N a}_{2} C O_{3} = 0.064$
$\frac{w}{106 / 2} \times 1000 = 0.064$
Mass of ${N a}_{2} C O_{3} = 3.392 \times 10^{- 3} g / 25 m L$
Strength of ${N a}_{2} C O_{3} = 0.1357 g / l i t r e$

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