Question

What mass of potassium fluoride must be added to 500.0 mL of water to give a solution with pH = 8.07?

[K_a(HF) = 7.1 × 10^–4]

		A. 68 g
		B. 1.7 × 10–4 g
		C. 1.96 g
		D. 0.95 g
		E. 2.91 g

Answer 1

let mass of KF = m gram
Molar mass of KF = 58.1 g/mol
number of moles of KF = mass/molar mass = m/58.1

Volume = 0.5 L

[KF] =number of moles/ volume
         = (m/58.1*0.5)
          = 0.0344 m

pH = -log [H+]
8.07 = -log [H+]
[H+] = 8.5*10^-9 M
[OH-] = 10^-14/ (8.5*10^-9) = 1.18*10^-6 M

F- + H2O ---> HF + OH-
                           x          x

Kb = 10^-14 /Ka
       = 10^-14/ (7.1*10^-4)
       = 1.41*10^-11

Kb = [HF] [OH]/ [F-]
1.41*10^-11 = (1.18*10^-6 )*(1.18*10^-6 ) / (0.0344m -1.18*10^-6 )
0.0344m -1.18*10^-6   = 10.11
m = 2943 g

Looks like I have done some calculation mistake somewhere
But procedure is same

Please try out and feel free to ask if you have any doubts