Find the pH of an equimolar solution of HCN and NH 4 OH- Given- K a HCN -6 - 10-10 K b NH 4 OH -1-8 - 10-5A- 8-45B- 4-74C- 12-5D- 9-24

PK_a= log(K_a) pK_a= log(6× 10^ 10)=(10 0.78) pK_a=9.22 pK_b= log(K_b) pK_b= log(1.8× 10^ 5)=(5 0.26) pK_b=4.74 For a salt of weak acid and weak base nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Salt of Weak Acid and Weak Base

Find the pH o...

Question

Find the $p H$ of an equimolar solution of $H C N$ and $N H_{4} O H$ .
Given: $K_{a} (H C N) = 6 \times 10^{- 10}$ $K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}$

8.45

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4.74

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12.5

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9.24

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Solution

The correct option is D 9.24
$p K_{a} = - log (K_{a}) p K_{a} = - log (6 \times 10^{- 10}) = (10 - 0.78) p K_{a} = 9.22$

$p K_{b} = - log (K_{b}) p K_{b} = - log (1.8 \times 10^{- 5}) = (5 - 0.26) p K_{b} = 4.74$

For a salt of weak acid and weak base pH is given as:
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) p H = 7 + \frac{1}{2} (9.22 - 4.74) p H = 7 + \frac{4.48}{2} p H = 9.24$

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