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 =
X2 = 1.4e-12
x = [OH-] = 1.18e-6
pOH = 5.93
pH = 14 - pOH
your answer for #2 is : 8.07
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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
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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
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