Calculating ph of titration between a weak acid and a strong base?

Question

A 0.035 ml solution of 0.2500 M HF is titrated with standardized 0.1532 M solution of NaOH at 25 Celsius. ( Ka = 6.8 * 10^(-4) )

I know the equivalence point is 57.11 mL. I'm trying to find

1) the pH at 0.50 mL before the equivalence point

2) the pH at the equivalence point

3) the pH at 0.50 ml after the equivalence point.

Help? I missed this question on a test and can't seem to work it out to work it out correctly.

Answer 1

if you know the equivalence point is 57.11 mL, then you are starting with 0.035 litres of 0.2500 Molar HF & not 0.035 ml

find moles of HF:

0.035 Litres @ 0.2500mol/litre = 0.00875 moles

2) @ the pH at the equivalence point those moles have been converted into F-1 & they are in (0.035 + 0.05711 = 0.09211 litres), which gives a molarity of fluoride of :

0.00875 moles / 0.09211 litres = 0.0950 Molar

the fluoride does a hydrolysis:

F-1  in water --> HF  & OH-

Khydrolysis = Kwater / Kacid = 1e-14 / 6.8e-4 = 1.47e-11

1.47 e-11 = [HF] [OH-] / [F-]

1.47 e-11 = [x] [x] / [0.0950]

X2 = 1.4e-12

x = [OH-] = 1.18e-6

pOH = 5.93

pH = 14 - pOH

your answer for #2 is : 8.07

================================

3) the pH at 0.50 ml after the equivalence point.

find the moles of excess NaOH:

0.00050 litres @ 0.1532 mol/litre = 7.66e-5 moles of NaOH

& these 7.66e-5 moles of excess OH- are in (0.09211 + 0.00050 = )0.09261 Litres,  which gives an OH- molarity of:

7.66e-5 / 0.09261 Litres = 8.27e=4 Molar OH-

8.27e=4 Molar OH- .... gives a pOH of 3.08... which gives a pH of 10.92

Your answer to # 3 is: pH = 10.92

======================================...

1) the pH at 0.50 mL before the equivalence point

is probably faster calculated with the Henderson–Hasselbalch equation, but

we have found that 0.05 ml of base has 7.66e-5 moles of OH- .....

so just before the equiv point , lets see what has happened to the original 0.00875 moles of HF

0.00875 moles  - 7.66e-5 = 0.008673 moles of Fluoride has been produced,

& only 7.66e-5 moles of HF are left

let's find the Molarities of these two in there 92.11 - 0.50 ml's:

0.008673 moles / 0.09161litres = 0.09468 Molar Fluoride

7.66e-5 moles / 0.09161 litres = 8.362e-4 Molar acid

Ka = [H+] [F-] / [HF]

6.8e-4 = [H+] [0.09468] / [8.362e-4]

[H+] = 6.00 e-6

Your answer for #1 : pH = 5.22