The hydrolysis constant of NH-4Cl is 5-6-10-6- Find the pH of nbsp- 0-1 molar solution of NH-4Cl at equilibrium- When log-1-79-0-253-3-1375-248-95

The correct option is A 3.13Given, K_h=5.6×10^ 6, [NH_4Cl]=0.1 M NH_4Cl is nbsp; a mixture of strong acid and weak base we know the relation nbsp; K_h=K_wK ...

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Standard XII

Chemistry

Salt of Strong Acid and Weak Base

The hydrolysi...

Question

The hydrolysis constant of $N H_{4} C l$ is $5.6 \times 10^{- 6}$ . Find the $p H$ of 0.1 molar solution of $N H_{4} C l$ at equilibrium. When $l o g 1.79 = 0.253$ .

3.13

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5.24

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8.95

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Solution

The correct option is A 3.13
Given, $K_{h} = 5.6 \times 10^{- 6}, [N H_{4} C l] = 0.1 M$
$N H_{4} C l$ is a mixture of strong acid and weak base
we know the relation
$K_{h} = \frac{K_{w}}{K_{b}}$
$5.6 \times 10^{- 6} = \frac{10^{- 14}}{K_{b}}$
$K_{b} = \frac{10^{- 14}}{5.6 \times 10^{- 6}}$
$K_{b} = 1.79 \times 10^{- 9}$
$p K_{b} = - l o g K_{b}$
$= - l o g 1.79 \times 10^{- 9}$
$= 9 - l o g 1.79$
$= 9 - 0.253$
$= 8.747$
Also, $p H = \frac{1}{2} \times (p K_{w} - p K_{b} - l o g c)$
$p H = \frac{1}{2} \times (14 - 8.75 - l o g 0.1)$
$p H = \frac{1}{2} \times (14 - 8.75 + 1)$
$p H = 3.125$

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Salt of Strong Acid and Weak Base

Standard XII Chemistry

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The hydrolysis constant of NH4Cl is 5.6×10−6. Find the pH of 0.1 molar solution of NH4Cl at equilibrium. When log 1.79=0.253.

The hydrolysis constant of $N H_{4} C l$ is $5.6 \times 10^{- 6}$ . Find the $p H$ of 0.1 molar solution of $N H_{4} C l$ at equilibrium. When $l o g 1.79 = 0.253$ .