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`        ka for ch3cooh is 1.8 find out the percentage dissociation of o.2 m ch3cooh in 0.1 m hcl`
11 months ago

Arun
24739 Points
```							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 =0.1MTotal [H+] concentration = x + 0.1Now, Ka for acetic acid = 1.8 x 10-5 1.8 x 10-5  = x (x + 0.1)/0.2on solving x = 0.35 * 10^-4 hence percentage dissociation = 0.35 * 10^-4 * 100 /0.2 = 0.0175  Hope it helps. Let us know if there is any issue.
```
11 months ago
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