Hydrofluoric acid is a weak aci with a K a =6.8 x 10 -4 . A titration of 25.0 ml
ID: 996014 • Letter: H
Question
Hydrofluoric acid is a weak aci with a Ka=6.8 x 10-4. A titration of 25.0 ml of 0.500 M hydrfluoric acid is carried out using 0.200 M NaOH. a) What is the initial pH of the hydrofluoric acid solution?
b) Write the chemical reaction equation of this titration.
c) 50.0 mL of .200 M NaOH is added to the intial HF solution. At what point in the titration does this represent (i.e., before, at, or after the equivalence oint)? Support answer with calculation and/or logical explanation.
d) Whiat is th pH of this solution after the 50.0 ml of .200 M NaOH is added to the intial HF solution. ( dont forrget to account for dilution)
Explanation / Answer
1)
a) initial pH :
HF --------------------> H+ + F-
0.50 0 0
0.50-x x x
Ka = [H+][F-]/[HF]
6.8 x 10^-4 = x^2 / 0.5-x
x^2 + 6.8 x 10^-4 x - 3.4 x 10^-4 = 0
x = 0.018
[H+] = 0.018 M
pH = -log [H+]
pH = 1.74
b)
HF + NaOH ------------------------> NaF + H2O
c)
millimoles of HF = 25 x 0.500 = 12.5
millimoles of NaOH = 50 x 0.2 = 10
millimoles of HF > millimoles of NaOH . so it is before equivalence point. so it is buffer solution
d)
HF + NaOH --------------> NaF + H2O
12.5 10 0 0
2.5 0 10 10
pH = pKa + log [NaF]/[HF]
pH = 3.17 + log (10/2.5)
pH = 3.77
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