Question

If the Ka of a monoprotic weak acid is 5.8 × 10-6, what is the pH of a 0.29 M solution of this acid?

Answer 1

Lets write the dissociation equation of CH3COOH

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

0.29 0 0

0.29-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 ((5.8*10^-6)*0.29) = 1.297*10^-3

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

so, x = 1.297*10^-3 M

So, [H+] = x = 1.297*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1.297*10^-3)

= 2.89

Answer: 2.89