pH of NH_4Cl solution is 5.28. Calculate the | vedclass

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(A) For a salt of a weak base and strong acid $(NH_4Cl)$,the $pH$ is given by the formula: $pH = 7 - \frac{1}{2}(pK_b + \log C)$.Given $pH = 5.28$ and

Vedclass · Accepted Answer

$2.624 	imes 10^{-4}$

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(A) For a salt of a weak base and strong acid $(NH_4Cl)$,the $pH$ is given by the formula: $pH = 7 - \frac{1}{2}(pK_b + \log C)$.Given $pH = 5.28$ and $C = 0.02 \ M$.$5.28 = 7 - 0.5(pK_b + \log(0.02))$.$1.72 = 0.5(pK_b - 1.699)$.$3.44 = pK_b - 1.699 \implies pK_b = 5.139$.$K_b = 10^{-5.139} \approx 7.26 	imes 10^{-6}$.The degree of hydrolysis $(h)$ is given by $h = \sqrt{\frac{K_h}{C}}$,where $K_h = \frac{K_w}{K_b}$.$K_h = \frac{10^{-14}}{7.26 	imes 10^{-6}} \approx 1.377 	imes 10^{-9}$.$h = \sqrt{\frac{1.377 	imes 10^{-9}}{0.02}} = \sqrt{6.885 	imes 10^{-8}} \approx 2.624 	imes 10^{-4}$.

$pH$ of $NH_4Cl$ solution is $5.28$. Calculate the degree of hydrolysis $(h)$ of its $0.02 \ M$ solution.

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