Question

The xd for 10-6 M CH3COOH with K = 1.8 x 10-5
(1) 4.24
(2) 424
(3) 1
(4) none of these​

Answer 1

Answer:

First we should write the reaction,

For the HCl→H

+

+Cl

−

after its complete dissociation

[Here neglected 10

−7

M of H

+

due to water dissociation because 0.1M is more than 100 times than 10

−7

M]

Now for the weak acid which not dissociate completely and always form an equilibrium:

CH

3

COOH+H

2

O=CH

3

COO

−

+H

+

0.01 0 0.1(from dissociation of HCl)initially

0.01−x x x+0.1 (after dissociation of acetic acid)

Now writing

Ka=[CH

3

COO

−

][H

+

]/[CH

3

COOH]

1.8×10

−5

=x×(0.1+x)/0.01

Approximation:the acid is weak and soassuming x<<0.1M,0.1+x 0.1M

0.1x=1.8×0.01×10

−5

M

so x=1.8×10

−6

M

so % ionisation of acid=x/0.01×100=1.8×10

−2

%=0.018%(very less due to common ion effect)

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