The pH of a 0.02 \; M\; NH_{4}Cl solution will be [ given K_{b}\left ( NH_{4}OH \right )=10^{-5} and log\; 2=0.301]

  • Option 1)

    2.65

  • Option 2)

    4.35

  • Option 3)

    4.65

  • Option 4)

    5.35

Answers (1)

p^{H}=7-\frac{1}{2}\; pK-\frac{1}{2}\; log\; c

       =7-\frac{5}{2}-\frac{1}{2}log\left ( 2\times 10^{-2} \right )

       =5.35


Option 1)

2.65

Option 2)

4.35

Option 3)

4.65

Option 4)

5.35

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