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Introduction and Degree of Hydrolysis

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Question

Four solutions of ${N H}_{4} C l$ are taken with concentrations $1 M, 0.1 M, 0.01 M$ and $0.001 M$ . Their degree of hydrolysis are $h_{1}, h_{2}, h_{3}, h_{4}$ . What is the graduation of degree of hydrolysis?

h_{1} > h_{2} > h_{3} > h_{4}

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h_{1} = h_{2} = h_{3} = h_{4}

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h_{4} > h_{3} > h_{2} > h_{1}

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Solution

The correct option is C $h_{4} > h_{3} > h_{2} > h_{1}$
$N H_{4} C l ⇌ N H_{4}^{+} + C l^{-}$
Degree hydrolysis, $h = \sqrt{\frac{K_{a}}{C}}$ ; where $K_{a}$ and $C$ are dissociation constant and concentration respectively.
Therefore, $h \propto \frac{1}{\sqrt{C}}$
Since the degree of hydrolysis of $1 M, 0.1 M, 0.01 M$ and $0.001 M$ are $h_{1}, h_{2}, h_{3}$ and $h_{4}$ respectively.
Therefore, $h_{4} > h_{3} > h_{2} > h_{1}$ .

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Similar questions

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Four solutions of NH4Cl are taken with concentrations 1M,0.1M,0.01M and 0.001M. Their degree of hydrolysis are h1,h2,h3,h4. What is the graduation of degree of hydrolysis?

The correct option is C h4>h3>h2>h1NH4Cl⇌NH+4+Cl−Degree hydrolysis, h=√KaC ; where Ka and C are dissociation constant and concentration respectively.Therefore, h∝1√CSince the degree of hydrolysis of 1M,0.1M,0.01M and 0.001M are h1,h2,h3 and h4 respectively.Therefore, h4>h3>h2>h1 .

Four solutions of ${N H}_{4} C l$ are taken with concentrations $1 M, 0.1 M, 0.01 M$ and $0.001 M$ . Their degree of hydrolysis are $h_{1}, h_{2}, h_{3}, h_{4}$ . What is the graduation of degree of hydrolysis?