Find the pH of a 500 mL solution containing 2-5 moles of KCN and 2-5 moles of HCN at 25- C- Given-The pK b of cyanide ion CN -is 4-7A- 9-3B- 7-3C- 10-3D- 8-3

Milli moles of nbsp; salt KCN = 2.5 × 500 = 1250 nbsp; Milli moles of HCN = 2.5 × 500 =1250 Therefore in the buffer solution [Anion of Salt] = [Acid] So, th ...

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Standard XII

Chemistry

Acidic Buffer Action

Find the pH o...

Question

Find the $p H$ of a $500 m L$ solution containing 2.5 moles of $K C N$ and 2.5 moles of $H C N$ at $25^{o} C$ .
Given:The $p K_{b}$ of cyanide ion $(C N^{-})$ is 4.7

9.3

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Solution

The correct option is A 9.3
Milli-moles of salt $K C N = 2.5 \times 500 = 1250$
Milli-moles of $H C N = 2.5 \times 500 = 1250$
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.
$\frac{[Anion of Salt]}{[Acid]} = 1$
$p H = p K_{a} + {log}_{10} (\frac{[Anion of Salt]}{[Acid]})$
or, $p H = p K_{a}$
$p K_{a} + p K_{b} = 14$ at $25^{o} C$
$p K_{a} o f H C N = 14 - p K_{b} = 14 - 4.7 = 9.3 p H = p K_{a} = 9.3$

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Acidic Buffer Action

Standard XII Chemistry

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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

Find the $p H$ of a $500 m L$ solution containing 2.5 moles of $K C N$ and 2.5 moles of $H C N$ at $25^{o} C$ .
Given:The $p K_{b}$ of cyanide ion $(C N^{-})$ is 4.7