Question

Calculate the pH of the solution that results from each of the following mixtures.

140.0mL of 0.26M HF with 220.0mL of 0.30M NaF

Answer 1

[HF] = 0.26 M

[NaF] = 0.30 M

[F-] = [NaF] = 0.30 M because 1F- is given by 1 NaF

The reaction we should be using is

HF    +     H2O <-----> F-     +      H3O+    because we can use it to calculate [H3O+] directly

0.26           -                0.30            0      (Initial)

- x              -                 + x              +x    (Change)

0.26 - x      -               0.3 + x           x     (Final)

Ka = x * (0.3 + x) / (0.26 - x)

Now since x is very small (0.3+x) = 0.3 and (0.26-x) = 0.26

Ka of HF = 7.1 x 10^-4

7.1 x 10^-4 = x * 0.3 / 0.26

x = 7.1 x 10^-4 * 0.26 / 0.3

x = 6.153 x 10^-4 M

So,

[H3O+] = 6.153 x 10^-4 M

pH = -log(6.153 x 10^-4)

pH = 3.21