Find the percentage ionisation of 0.2M CHCOOH solution, whose dissociation constant is 1.8x10-5. Also
determine the concentration of H,O* and CH COOH at equilibrium
Answers
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)
Explanation: please vote me