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pH the Power of H

A hand book s...

Question

A hand book states that the solubility of $R N H_{2} (g)$ in water at $1$ atm and $0^{o}$ C is $22.41$ volumes of $R N H_{2} (g)$ per volume of water. ( $p K_{b}$ of $R N H_{2} = 4$ ) Find the max. pOH that can be attained by dissolving $R N H_{2}$ in water.

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Solution

The correct option is C $2$
$n_{R N H_{2} (g)} = \frac{1 \times 22.41}{0.0821 \times 273} = 1$
$[R N H_{2} (a q .)] = \frac{n_{R N H_{2} (g)}}{Volume of solution} = 1$
$α$ is negligible w.r.t. 1 so
$p O H = \frac{1}{2} [p K_{b} - l o g C] = 2$

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