Hydrofluoric acid is a weak acid with a Ka = 6.8 x 10-4 . A titration of 25.0 mL
ID: 996109 • Letter: H
Question
Hydrofluoric acid is a weak acid with a Ka = 6.8 x 10-4 . A titration of 25.0 mL of 0.500 M hydrofluoric 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 for this titration. c. 50.0 mL of 0.200 M NaOH is added to the initial HF solution. At what point in the titration does this represent (i.e., before, at, or after the equivalence point)? Support your answer with calculations and/or a logical explanation. d. What is the pH of this solution after the 50.0 mL of 0.200 M NaOH is added to the initial HF solution? (Do not use Henderon Hasselbach Equation)
Explanation / Answer
a ) intial concentration of H+ = (Ka * C).5 = 3.4*10-4
pH = -log [ H+] = 3.47
b. HF + NaOH = NaF + H2O
c.50 ml (V1) NaOH of strenght S1 = .2 M is trited aganist .5 M S2 HF
V1 S1 = V2 S2
V2 =20 ml
25 ml - 20 ml = 5 ml HF is remaining after adding 50 ml NaOH so it is before equivalence point.
d) now after adding 50 ml NaOH total volume of system became 75 ml, 5 ml HF of concentration .5 M is remaining so it concentration will reduce . which will be .03 M
H+ = (Ka * C).5 = 2.27 *10-4
pH = 4.7
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