$2 g N a O H$ is added to $100 m l (\frac{M}{20}) H_{2} S O_{4}$ solution and the resulting solution is diluted by the addition of $900 m l H_{2} O$ .The pH of the solution will be?(round off to the next nearest integer) $\log 4 = 0.6$

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answer is 13.

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Detailed Solution

M. eq. of NaOH taken $= \frac{2}{40} \times 1000 = 50$
M eq. of $H_{2} S O_{4}$ taken $= 100 \times \frac{1}{20} \times 2 = 10$
$[O H^{-}] = \frac{50 - 10}{1000} = \frac{40}{1000} = 4 \times 10^{- 2}$

$p O H = 2 - \log 4 = 2 - 0.6 = 1.4$

$∴ p H = 14 - p O H = 14 - 1.4 = 12.6$

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