Question

The pH of an aqueous solution of 0.441 M acetic acid is​ _______________

Answer 1

Ka of CH3COOH = 1.8*10^-5

Lets write the dissociation equation of CH3COOH

CH3COOH -----> H+ + CH3COO-

0.441 0 0

0.441-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.441) = 2.817*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.817*10^-3 M

So, [H+] = x = 2.817*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.817*10^-3)

= 2.55

Answer: 2.55