Question

Hydrofluoric acid, HF is a weak acis with Ka= 3.5X10^-4. It dissociates according to

HF(aq)    H^+(aq) + F^-(aq)

a. What is the pH of a buffer solution which is 0.30M in F- and 0.60M in HF?

b. What will be the pH of the resulting solution when 2.0mL of 0.5M HCl are added to 50mL of this (from part a) buffer solution

c. What will be the pH of the resulting solution when 2.0mL of 0.5M NaOH are added to 50mL of this (from part a) buffer solution?

Answer 1

1.

Thus,




2.
No. of moles of H+ in 50mL of sol. from part a= 7 x 10^-4 x (50/1000) = 3.5 x 10^-5 moles
No. of moles of H+ in 2 mL of 0.5M HCl sol. = 0.5 x (2/1000) = 10^-3 moles
Total number of moles of H+ in resulting sol. = 3.5 x 10^-5 + 10^-3 = 0.001035 moles in 50+2=52 mL solution

Thus, [H+] = (0.001035/52) x 1000 = 0.0199 M
Thus, pH = -log(0.0199) = 1.7

3.
No. of moles of NaOH in 2mL of 0.5 solution = 0.5 x (2/1000) =10^-3 moles
And 3.5 x 10^-5 moles of H+ present in the solution will react with excess of NaOH,
Thus, number of moles of unreacted OH- ions in solution are,
10^-3 - 3.5 x 10^-5 = 0.000965 moles of OH- in 52 mL solution
[OH-]=0.000965 x (1000/52) = 0.01856 M
pOH= -log[OH-] = 1.73
pH = 14 - pOH = 12.27