what is the pH of a .01 M solution of potassium formate, HCOOK?

Question

the pK_a of formic acid, HCOOH, is 3.75.

Answer 1

Potassium format is the salt formed from a strong base (KOH) and a weak acid (formic acid).  Thus, the pH of a resulting solution will be >7 (alkaline).  Here is how you approach such a problem.

 

HCOOK will hydrolyze in water:  HCOOK + H₂O ==> HCOOH + KOH ... eq. 1

Looked at another way, HCOO- + H₂O ===> HCOOH + OH-  ... eq 2  (HCOO^- is the conjugate base of HCOOH)

 

Since HCOO^- is a base, you can use Kb = [HCOOH][OH^-]/[HCOO^-]  from eq. 2

pK_a = 3.75, therefore Ka = 1x10^-3.75 = 1.78x10^-4

Kb = 1.78x10-4 = (x)(x)/0.01-x  and assuming x is small relative to 0.01, we can ignore it (for now)

1.78x10^-4 = x²/0.01

x2 = 1.78x10^-6

x = 1.33x10^-3 (note: this value is not small relative to 0.01, so the quadratic should be used)

Using the quadratic...

x² + 1.78x10^-4 -1.78x10^-6

x = 0.001248 = [OH^-]

pOH = -log 0.001248 = 2.9

pH = 14 - 2.9 = 11.1