The pH of a 0.02 M NH_4Cl solution will be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.The formula for the $pH$ of such a salt solution is:$pH = \frac{1}{2} [pK_w

Vedclass · Accepted Answer

$5.35$

Vedclass · Answer

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.The formula for the $pH$ of such a salt solution is:$pH = \frac{1}{2} [pK_w - pK_b - \log C]$Given:$C = 0.02 \ M = 2 	imes 10^{-2} \ M$$K_b = 10^{-5} \implies pK_b = 5$$K_w = 10^{-14} \implies pK_w = 14$Substituting the values:$pH = \frac{1}{2} [14 - 5 - \log(2 	imes 10^{-2})]$$pH = \frac{1}{2} [9 - (\log 2 + \log 10^{-2})]$$pH = \frac{1}{2} [9 - (0.301 - 2)]$$pH = \frac{1}{2} [9 - (-1.699)]$$pH = \frac{1}{2} [10.699]$$pH = 5.3495 \approx 5.35$

The $pH$ of a $0.02 \ M \ NH_4Cl$ solution will be [given $K_b \ (NH_4OH) = 10^{-5}$ and $\log \ 2 = 0.301$ ]

Similar Questions

The aqueous solution of $AlCl_3$ is:

$A$ solution of sodium borate has a $pH$ of approximately

$A$ $100 \ L$ solution of $KCN$ undergoes $3.7\%$ hydrolysis. If the dissociation constant of $HCN$ is $7.2 \times 10^{-10}$,find the value of $K_w$.

At $25 \ ^\circ C$,a weak acid $HX$ has the dissociation constant $K_a = 1 \times 10^{-5}$. It forms a salt $NaX$ on reaction with an alkali. The percentage of degree of hydrolysis of a $0.1 \ M$ solution of $NaX$ is $..... \%$.

Calculate the hydrolysis constant $(K_h)$ for a salt containing $NO_2^-$. Given that the dissociation constant $(K_a)$ for $HNO_2$ is $4.5 \times 10^{-4}$ and the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module