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  4. Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924 K. The p Ka value of HA is [. Given = molarity = molality, [ mathrmK mathrmf( mathrmH2 mathrmO)=1.85 mathrmKkg mathrmmol-1, log 2=0.3]

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Q. Depression in freezing point of $0.01$ molal aqueous solution of $HA$ is $0.01924\, K$. The $p K_{a}$ value of $HA$ is
$\left[\right.$ Given $=$ molarity $=$ molality, $\left[\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right)=1.85 \mathrm{Kkg} \mathrm{mol}^{-1}, \log 2=0.3\right]$

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Solution:

$\Delta T_{f}=i \times m \times K_{f}$
$0.01924= i \times 0.01\times 1.85$
$i =1.04$
$i =1+\alpha( n -1)$
$1.04=1+\alpha(2-1)$
$\alpha=0.04$
$\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}=(0.04)^2 \times 0.01=16 \times 10^{-6}$
$pK _{ a }=4.8$