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Question

# Ka for CH3COOH is 1.8 × 10−5. 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 0.018The dissociation of acetic acid can be represented as:CH3COOH↔H++CH3COO−HCl dissociated completely.Now, let the concentration of acetate ion be x.Therfore, hydrogen ion contribution from acetic acid=xSince HCL dissociated completely, hydrogen ion contribution from HCl=0.1 MTotal [H+] concetration=x+0.1Now, Ka for acetic acid=1.8×10−5Hence,Ka=[CH3COO−][H+]CH3COOH]1.8×10−5=x(x+0.1)0.20.36×10−5=x(x+0.1)x2+0.1x−0.36×10−5x=0.35×10−4Percentage dissociation=0.35×10−4×1000.2=0.0175≈0.018

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