Find the pH at each of the following points in the titration of 25 ml of 3.0M HF with 0.3 M NaOH. The Ka value is 6.6x10-4. After adding 25 ml of 0.3M NaOH

Answer 1

HF + NaOH ==> NaF + H2O

millimols HF initially = mL x M = 25 x 3 = 75
mmols NaOH added = 25 x 0.3 = 7.5
mmols HF left un-neutralized = 75-7.5 = 67.5
volume of new solution = 26 mL + 25 mL = 50 mL.
M HF left = mmol/mL = 67.5/50 = 1.35 M.

......HF ==> H^+ + F^-
I....1.35.....0.....0
C.....-x......x.....x
E....1.5-x....x.....x

Substitute the E line into Ka expression for HF and solve for x = (H^+), then convert to pH.