Calculate pH of 0.002 N NH4OH having 2% dissociation

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7.6

8.6

9.6

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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