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Question

Find the pH of a 500 mL solution containing 2.5 moles of KCN and 2.5 moles of HCN at 25oC.
Given:The pKb of cyanide ion (CN) is 4.7

A
9.3
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B
7.3
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C
10.3
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D
8.3
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Solution

The correct option is A 9.3
Milli-moles of salt KCN=2.5×500=1250
Milli-moles of HCN=2.5×500=1250
Therefore in the buffer solution [Anion of Salt] = [Acid]
So, this quantity is the ratio of concentration of the anion of the salt and the acid present in the mixture will be equal to 1.
[Anion of Salt][Acid]=1
pH=pKa+log10([Anion of Salt][Acid])
or, pH=pKa
pKa+pKb=14 at 25oC
pKa of HCN=14pKb=144.7=9.3pH=pKa=9.3

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