pKb of aq. NH3 is 4.74, hence pH of 0.01 M NH3 solution is:

pOH=12pKb−log C pOH=124.74-12log 0.01=3.37 pH=14-pOH pH=14-3.37=10.63

pKb of aq. NH3 is 4.74, hence pH of 0.01 M NH3 solution is:

pOH=12pKb−log C pOH=124.74-12log 0.01=3.37 pH=14-pOH pH=14-3.37=10.63

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pK_b of aq. NH₃ is 4.74, hence pH of 0.01 M NH₃ solution is:

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12.00

2.00

10.63

3.37

answer is B.

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Detailed Solution

$pOH = \frac{1}{2} [{pK}_{b} - \log C] pOH = \frac{1}{2} (4.74) - \frac{1}{2} \log [0.01] = 3.37 pH = 14 - pOH pH = 14 - 3.37 = 10.63$

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pKb of aq. NH3 is 4.74, hence pH of 0.01 M NH3 solution is:

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pK_b of aq. NH₃ is 4.74, hence pH of 0.01 M NH₃ solution is: