A monoprotic acid in a 0.1,,M solution ionize | vedclass

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

Question

(d) $	herefore$ Monoprotic acid $HA$

$HA$ $ ightleftharpoons $ ${H^ + } + {A^ - }$

Ionisation constant = ?

$\alpha = 0.001\,\% = \frac{{0.

Vedclass · Accepted Answer

$1.0 	imes {10^{ - 11}}$

Vedclass · Answer

(d) $	herefore$ Monoprotic acid $HA$

$HA$ $ ightleftharpoons $ ${H^ + } + {A^ - }$

Ionisation constant = ?

$\alpha = 0.001\,\% = \frac{{0.001}}{{100}} = {10^{ - 5}}$

$K = \frac{{{\alpha ^2}}}{V} = \frac{{{{[{{10}^{ - 5}}]}^2}}}{{10}} = {10^{ - 11}}$.

A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is

Similar Questions

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

A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$

${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.

A weak base $MOH$ of $0.1\,N$ concentration shows a $pH$ value of $9$ . What is the percentage degree of ionization of the base ? .......$\%$

The $pH$ of a $0.1\ M$ aqueous solution of a very weak acid $(HA)$ is $3$. What is its degree of dissociation ?......$\%$