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Q.

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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