Which of the following formulae is used to fi | vedclass

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

Question

(B) For a weak monoacidic base $BOH$,the dissociation equilibrium is represented as: $BOH \rightleftharpoons B^{+} + OH^{-}$.Let the initial concentra

Vedclass · Accepted Answer

$\sqrt{K_{b} \cdot c}$

Vedclass · Answer

(B) For a weak monoacidic base $BOH$,the dissociation equilibrium is represented as: $BOH \rightleftharpoons B^{+} + OH^{-}$.Let the initial concentration be $c$ and the degree of dissociation be $\alpha$.The equilibrium concentrations are: $[BOH] = c(1-\alpha)$,$[B^{+}] = c\alpha$,and $[OH^{-}] = c\alpha$.The dissociation constant $K_b$ is given by: $K_b = \frac{[B^{+}][OH^{-}]}{[BOH]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$.Since the base is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$.Thus,$K_b \approx c\alpha^2$,which gives $\alpha = \sqrt{\frac{K_b}{c}}$.The concentration of $[OH^{-}]$ is $c\alpha = c \cdot \sqrt{\frac{K_b}{c}} = \sqrt{K_b \cdot c}$.

Which of the following formulae is used to find the $[OH^{-}]$ ion concentration of a weak monoacidic base?

Similar Questions

The dissociation constant of a monoacidic weak base $MOH$ is $1.8 \times 10^{-5}$. What is the concentration of $OH^{-}$ ions in its $0.1 \, M$ solution?

The ionization constant of $0.1$ $M$ weak acid is $1.74 \times 10^{-5}$ at $298$ $K$ temperature. Calculate the $pH$ of its $0.1$ $M$ solution. (in $.88$)

Find the $pOH$ value of $10^{-3} \ M \ NH_4OH$. Given $K_b = 10^{-5}$.

In its $0.2 \, M$ solution,an acid ionises to an extent of $60 \%$. Its hydrogen ion concentration is.....$M$

The $pH$ of $0.1 \ M$ $HCN$ solution is $5.2$. Calculate the dissociation constant $K_a$ of this solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module