Question

A 100.0 mL sample of 0.20 M HF is titrated with 0.20 M KOH. Determine the pH of the solution after the addition of 50.0 mL of KOH. The Ka of HF is 3.5 × 10-4.

A. 2.08

B. 3.15

C. 4.33

D. 3.46

E. 4.15

Answer 1

answer : D) 3.46

millimoles of HF = 100 x 0.2 = 20

millimoles of KOH = 50 x 0.20 = 10

Ka = 3.5 x 10^-4

pKa of HF = 3.46

HF    +   KOH    -----------------> KF   + H2O

20          10                                  0         0

10             0                                10         10

this is half equivalence point. so pH = pKa

pH = 3.46