Calculate the ph of a 3.78*10-3 m solution of naf, given that the ka of hf = 6.80 x 10-4 at 25°c.

The answer is pH = 7.37.

Solution:

First, we set up an ICE table for the reaction:

F - + H2O → HF + OH-

Initial 0.00378 0 0

Change - x + x + x

Equilibrium 0.00378-x x x

We can calculate Kb from the given Ka, since we know that Kw = Ka*Kb = 1.0 x 10^-14 at 25°C:

Kb = Kw/Ka = 1.0x10^-14 / 6.80x10^-4 = 1.471 x 10^-11

Kb = 1.471 x 10^-11 = [OH-][HF] / [F-] = (x) (x) / (0.00378-x)

Approximating that x is negligible compared to 0.00378 simplifies the equation to

1.471x10^-11 = (x) (x) / 0.00378

1.471 x 10-11 = x2 / 0.00378

Then we solve for x that is also equal to [OH-]:

x2 = (1.471 x 10^-11) (0.00378)

x = sqrt[ (1.471 x 10^-11) (0.00378) ] = 2.358x10^-7 = [OH-]

in which 0.0000002358 is indeed negligible compared to 0.00378.

We can now calculate for pOH:

pOH = - log [OH-] = - log (2.358x10^-7) = 6.63

Therefore, the pH is

pH = 14 - pOH = 14 - 6.63 = 7.37