Chemistry · 2019
JEE Main · 10 April 2019, Shift 2 · Q57
The pH of a 0.02 M NH_4 Cl solution will be [given K_b ( NH_4 OH )=10^-5 and log 2=0.301 ]
The pH of a $\displaystyle 0.02 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ solution will be [given $\displaystyle \mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=10^{-5}$ and $\displaystyle \log 2=0.301$ ]
Official answer
(3)
$\displaystyle 5.35$
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JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.