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

Question

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is (A) 1.0 X 10-3 (B) 1.0 X 10-6 (C) 1.0 X 10-8 (D) 1.0 X 10-11

Tardigrade Admin · Accepted Answer

Ionisation constant, Ka=(a2 × c/1-a) α= degree of dissociation = 0.001 % =(0.001/100)=1×10-5 C= concentration = 0.1 M when α is very small K=α2 × C =(1 × 10-5)2 × 0.1=1× 10-11

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

$1.0 X 1 0^{- 3}$

$1.0 X 1 0^{- 6}$

$1.0 X 1 0^{- 8}$

$1.0 X 1 0^{- 11}$

Ionisation constant, $K_{a} = \frac{a ^{2} \times c}{1 - a}$
$α$ = degree of dissociation $= 0.001%$
$= \frac{0.001}{100} = 1 \times 1 0^{- 5}$
$C$ = concentration $= 0.1 M$
when $α$ is very small
$K = α^{2} \times C$
$= (1 \times 1 0^{- 5})^{2} \times 0.1 = 1 \times 1 0^{- 11}$

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

1.0X10−3

1.0X10−6

1.0X10−8

1.0X10−11

Ionisation constant, Ka​=1−aa2×c​ α= degree of dissociation =0.001% =1000.001​=1×10−5 C= concentration =0.1M when α is very small K=α2×C =(1×10−5)2×0.1=1×10−11

$1.0 X 1 0^{- 3}$

$1.0 X 1 0^{- 6}$

$1.0 X 1 0^{- 8}$

$1.0 X 1 0^{- 11}$

Ionisation constant, $K_{a} = \frac{a ^{2} \times c}{1 - a}$
$α$ = degree of dissociation $= 0.001%$
$= \frac{0.001}{100} = 1 \times 1 0^{- 5}$
$C$ = concentration $= 0.1 M$
when $α$ is very small
$K = α^{2} \times C$
$= (1 \times 1 0^{- 5})^{2} \times 0.1 = 1 \times 1 0^{- 11}$