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Q.

Degree of hydrolysis of 0.1MCH3COONH4,When KaCH3COOH=KbNH4OH=2.0×10-5 is

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a

0.5×10-2

b

0.93

c

0.75×10-2

d

7.63×10-3

answer is A.

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Detailed Solution

Kh=KwKa×Kb =10-142×10-5×2×10-5=0.5×10-2
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