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Calculate pH of 0.002 N NH4OH having 2% dissociation
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a
7.6
b
8.6
c
9.6
d
10.6
answer is C.
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Detailed Solution
NH4OH is a weak base and partially dissociated NH4OH ⇌ NH4+ + OH-Concentration 1 0 0before dissociationConcentration 1-α α αafter dissociationC=2×10-3 ; α=2100 ∴ OH- = Cα = 2 × 10-3 × 2100 = 4 × 10-5 MpOH = -log OH- = -log 4 × 10-5 = 4.4 pH=14-4.4 = 9.6
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