JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 6)
The pH of a 0.02 M NH4Cl solution will be :
[given Kb (NH4OH) = 10–5 and log 2 = 0.301]
[given Kb (NH4OH) = 10–5 and log 2 = 0.301]
2.56
5.35
4.35
4.65
Explanation
NH4+ + H2O ⇋ NH4OH + H+
[H+] = c$$\alpha $$
= $$\sqrt {{k_a}\left( {NH_4^ + } \right) \times c} $$
= $$\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c} $$
= $$\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02} $$
= $$\sqrt {20} \times {10^{ - 6}}$$
$$ \therefore $$ pH = $$ - \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)$$ = 5.35
[H+] = c$$\alpha $$
= $$\sqrt {{k_a}\left( {NH_4^ + } \right) \times c} $$
= $$\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c} $$
= $$\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02} $$
= $$\sqrt {20} \times {10^{ - 6}}$$
$$ \therefore $$ pH = $$ - \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)$$ = 5.35
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