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  4. The pH of a 0.02M NH4Cl solution will be [given Kb(NH4OH) = 10-5 and log2=0.301]

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Q. The pH of a 0.02M $NH_4Cl$ solution will be [given $K_b(NH_4OH) = 10^{-5}$ and log2=0.301]

JEE MainJEE Main 2019Equilibrium

Solution:

For the salt of strong acid and weak base
$H^{+} = \sqrt{\frac{K_{W} \times C}{K_{b}}} $
$ \left[H^{+}\right] = \sqrt{\frac{10^{-14} \times2 \times10^{-2}}{10^{-5}} }$
$- \log \left[H^{+}\right] = 6 - \frac{1}{2} \log 20 $
$ \therefore pH = 5.35 $