The degree of hydrolysis of a 0.1 M solution | vedclass

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(C) For the salt of a weak base and a strong acid,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}

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$0.005$

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(C) For the salt of a weak base and a strong acid,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_b \cdot C}}$.From this,we can see that $h \propto \frac{1}{\sqrt{C}}$,which implies $\frac{h_1}{h_2} = \sqrt{\frac{C_2}{C_1}}$.Given: $C_1 = 0.1 \ M$,$h_1 = 0.01$,and $C_2 = 0.4 \ M$.Substituting the values: $\frac{0.01}{h_2} = \sqrt{\frac{0.4}{0.1}} = \sqrt{4} = 2$.Therefore,$h_2 = \frac{0.01}{2} = 0.005$.

The degree of hydrolysis of a $0.1 \ M$ solution of $ArNH_3^+ Cl^-$ is $0.01$. Find the degree of hydrolysis in a $0.4 \ M$ solution of $ArNH_3^+ Cl^-$.

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