Question

Given a 1.00 L solution that is 0.60 M HF and 1.00 M KF, calculate the pH after 0.080 mol NaOH is added, and calculate the pH after 0.24 mol HCl is added to the original solution.

Answer 1

0.60M HF/1.00M KF Buffer Soln
Before adding OH- or H+,
Ka=[H+][F-]/[HF]
=> [H+]=Ka[HF]/[F-]
=> [H+]=4.2E-4(1.00)/(0.60)
=> [H+]=0.0007M in H+
=> pH=-log(0.0007)=3.15

HF => H+ + F-
0.60 x 1.00
Adding 0.08M NaOH Shifts Equilibrium Right, decreasing HF by 0.08M and increasing F- by 0.08M.

New Concentrations
HF => H+ + F-
0.52M x 1.08M

You'll need Ka for HF to do the calculation (Ka(HF) = 7.2E-4)This can be found in most Ka tables for weak acids.

7.2E-4 = x(1.08)/0.52; solve for x = H+ in new equilibrium
= [7.2E-4(0.52)/1.08]M
= 0.000347M
pH = -log(0.000347) = 3.46
NOTE: Adding OH- always shifts buffer pH to a higher pH.

Adding 0.24 HCl shifts the equilibrium left; increasing HF by 0.24M and decreasing F- by 0.24M.

New Equilibrium Concentrations:
HF => H+ + F-
(0.60+0.24) x (1.00-0.24)
=0.84M =0.76M

7.2E-4 = x(0.76)/0.84
=> x = [7.2E-4(0.84)/0.76] = 0.0008M in H+
pH = -log(0.0008)= 3.1
NOTE: Adding an acid shifts the pH to a lower pH value.