Ka for CH,COOH is 1.8 x 10-5. Find out the
percentage dissociation of 0.2M CH,COOH in
0.1M HCl solution
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The dissociation of acetic acid can be represented as:
CH3COOH ↔H+ + CH3COO−
HCl dissociated completely.
Now, let the concentration of acetate ion be x. Therefore, hydrogen ion contribution from acetic acid = x.
Since HCl dissociates completely, hydrogen ion contribution from HCl =0.1M
Total [H+] concentration = x + 0.1
Now, Ka for acetic acid = 1.8 x 10-5
Hence,
Ka=[CH3COO−][H+][CH3COOH]1.8×10−5=(x)×(x+0.1)0.20.36×10−5 = (x)×(x+0.1)x2+0.1x−0.36×10−5=0x=0.35 × 10−4
Percentage dissociation = 0.35 × 10−4×1000.2=0.0175
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