Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.

A. a mixture that is 0.15M in HF and 0.15M in NaF
B.0.15M NaF
C. 0.15M HF

Answer 1

1.
This is a simple weak acid problem and can use this equation:
No Ka was provided so I'm going to use one online.
Ka=[H⁺][F^-]/[HF]
7.2x10^-4=[x²]/.15
=.0104 M
Solve for x which will give the concentration of H⁺
Now just do this:
pH=-log(x)
=1.98

2.
Now the F^- ion is the conjugate base of HF so we wouldexpect to have a pH greater than 7 and therefore we should convertthe Ka to Kb using this:

1x10^-14=(7.2x10^-4)Kb
Kb=1.39x10^-11

Since its a conjugate base, it will follow this reaction:

F^- + H₂O<-->HF+OH^-

Now we just need to set up an equilibrium expression:

1.39x10^-11=[HF][OH^-]/[F^-]
Following the same steps from number one, we should get the pOH tobe 5.84. Now this is pOH, we need pH, so just subtract this from 14to get:
8.16

3.
Now this is a buffer solution, we can still use the equilibriumexpression from 1. but with a few modifications.
Ka=[H⁺][F^-]/[HF]
We need to account for the extra F^- present, so we add.15 (from the NaF) to the F^- concentration:
Ka=[H⁺][.15+F^-]/[HF]
7.2x10^-4=[x][x+.15]/[.15]
Now x in this case is really really small and we can discount itfrom F^-
7.2x10^-4=[x][.15]/[.15]
Solve for x, then plug into -log(x) to get the pH which should be3.14