PH of 0.1 M acetic acid is 3 , dissociation constant of acid will be

English

Gujarati

Hindi

6-2.Equilibrium-II (Ionic Equilibrium)

medium

The $ pH$ of $ 0.1$ $M$ acetic acid is $3$, the dissociation constant of acid will be

A

$1.0 \times {10^{ - 4}}$

B

$1.0 \times {10^{ - 5}}$

C

$1.0 \times {10^{ - 3}}$

D

$1.0 \times {10^{ - 8}}$

Solution

(b) $pH = 3$, $[{H^ + }] = {10^{ – 3}}\,M$

$\because [{H^ + }] = \sqrt {K \times c} $

${[{10^{ – 3}}]^2} = K \times c$ ; $\frac{{[{{10}^{ – 6}}]}}{{0.1}} = K = {10^{ – 5}}$

Standard 11

Chemistry

Share

Similar Questions

The hydrogen ion concentration of a $0.006\,M$ benzoic acid solution is $({K_a} = 6 \times {10^{ – 5}})$

medium

Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?

medium

A solution of weak acid $HA$ containing $0.01$ moles of acid per litre of solutions has $pH = 4$. The percentage degree of ionisation of the acid and the ionisation constant of acid are respectively.

medium

Dissociation constant for a monobasic acid is $10^{-4}$ . What is the $pH$ of the monobasic acid ? (If $\%$ dissociation $= 2\,\%$ )

medium

The ionization constant of propanoic acid is $1.32 \times 10^{-5}$. Calculate the degree of ionization of the acid in its $0.05\, M$ solution and also its $pH$. What will be its degree of ionization if the solution is $0.01$ $M$ in $HCl$ also?

medium