Question

Propanoic acid (HC3H8O2) has a Ka of 1.3*10^-5. if you are titrating a 25.00 mL sample of 0.355 M propanoic acid with 0.525 M NaOH, calculate the pH before the titration has started.

Answer 1

HC3H8O2 dissociates as:

HC3H8O2 -----> H+ + C3H8O2-

0.355 0 0

0.355-x x x

Ka = [H+][C3H8O2-]/[HC3H8O2]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.3*10^-5)*0.355) = 2.148*10^-3

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

so, x = 2.148*10^-3 M

So, [H+] = x = 2.148*10^-3 M

use:

pH = -log [H+]

= -log (2.148*10^-3)

= 2.6679

Answer: 2.67