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Question

Ka for CH3COOH is 1.8 × 105. Find out the percentage dissociation of 0.2 M CH3COOH in 0.1 M HCl solution ________.

A
0.018
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B
0.36
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C
18
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D
36
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Solution

The correct option is A 0.018
The dissociation of acetic acid can be represented as:

CH3COOHH++CH3COO

HCl dissociated completely.
Now, let the concentration of acetate ion be x.
Therfore, hydrogen ion contribution from acetic acid=x
Since HCL dissociated completely, hydrogen ion contribution from HCl=0.1 M
Total [H+] concetration=x+0.1
Now, Ka for acetic acid=1.8×105

Hence,
Ka=[CH3COO][H+]CH3COOH]

1.8×105=x(x+0.1)0.2

0.36×105=x(x+0.1)

x2+0.1x0.36×105

x=0.35×104

Percentage dissociation=0.35×104×1000.2=0.01750.018

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