Solve this following

Question:

The $\mathrm{pH}$ of a $0.02 \mathrm{M} \mathrm{NH}$ [given $\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}$ and $\log 2=0.301$ ]

  1. $4.65$

  2. $5.35$

  3. $4.35$

  4. $2.65$


Correct Option: , 2

Solution:

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