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Question

Calculate the degree of hydrolysis and pH of 0.02M ammonium cyanide (NH4CN) at 298 K. [K1 of HCN=4.99×109,Kb for NH4OH=1.77×105]

A
8.2
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B
3.2
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C
9.3
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D
3.9
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Solution

The correct option is D 9.3
Kh=10144.99×109×1.77×1015=1.132

It can be seen that hydrolysis constant (Kh) is not small, and for calculating h, the equation used is

h=Kh1+Kh=1.061+1.06=0.51

Using the formula
pH=pKalog h +log (1h)
=log(4.99×1010)log (0.51)+ log(10.51)
=9.3

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