Question

1. Valeric acid is a monoprotic acid with a K_a value of 1.44 x 10^-5. A student prepared 100.0 mL of a 0.200 M solution of valeric acid. What is the pH of the valeric acid solution?

Answer 1

Lets write the acid as HA

Lets write the dissociation equation of HA

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

0.2 0 0

0.2-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 ((1.44*10^-5)*0.2) = 1.697*10^-3

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

so, x = 1.697*10^-3 M

So, [H+] = x = 1.697*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1.697*10^-3)

= 2.77

Answer: 2.77