Question

Ph of 0.01m acetic acid with degree of dissociation 0.001 is

Answer 1

dissociation reaction of acetic acid is
$CH_3COOH\rightarrow CH_3COO^-+H^+$

so, concentration of [H^+] = $C\alpha$
where C is concentration of acetic acid and $\alpha$ is degree of dissociation.

so, [H^+] = 0.01 × 0.001 = 10^-5

so, pH = -log[H^+] = -log(10^-5) = 5

hence, pH is 7

Answer 2

pH= 1/2 pka + 1/2 log C

Ka=dissociation constant

C= concentration of acetic acid

degree of dissociation=α

α =√(Ka/C)

∴ Ka= 0.01*0.001*0.001 =10^-8

pH=1/2*(-log ka) - 0.01/2

    =4-0.005=3.995

Acetic acid very weak acid.It dissociates partially in aqueous solution.Its dissociation constant is very low as degree of dissociation is very small.

pH of 0.01m acetic acid with degree of dissociation 0.001 is 3.995.

acetic acid is required to prepare buffer.In vinegar there is acetic acid.It is also weak electrolyte.Its conductivity is very low.