The acid ionization (hydrolysis) constant of Zn2+ is 1.0 × 10-9. Hence, pH of 0.001 M solution of ZnCI2 is......

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answer is 6.

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Detailed Solution

Zn2+H2O⇌Zn(OH)++H+Kh=1×10−9Kh=KwKbZn(OH)2)∴ Kb=KwKh=1×10−5 pKb=5∴ pH=7−pKb+log⁡C2=7−(5−3)2=6

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