Question

The Ka of a citric acid is 1.95x10^-5 at 25 C. what is the pH of a 0.35 M aqueous solution of citric acid?

Answer 1

Lets write the citric acid as HA

Lets write the dissociation equation of HA

HA -----> H+ + A-

0.35 0 0

0.35-x x x

Ka = [H+][A-]/[HA]

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 ((1.95*10^-5)*0.35) = 2.612*10^-3

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

so, x = 2.612*10^-3 M

So, [H+] = x = 2.612*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.612*10^-3)

= 2.58

Answer: 2.58