Question

a solution that is made by combining 55 mL of 6.0×10−2 M hydrofluoric acid with 125 mL of 0.11 M sodium fluoride

Answer 1

Hydrogen Fluoride ------------> HF

HF -----------> H⁺ + F^-

Ka= [Product] / [Reactant]

    = [H⁺] [F^‑] / [HF]

   = 7.2X10^-4 ( This value can be directly substituted from Handbook)

After Mixing,

[HF] = (6X10^-2 X 55) / (125+55)

         = 0.01833M

[F^‑] = (0.11 X 125) / (125+65)

      = 0.07638M

Let the amount of HF that ionizes for the dissociation is s

[HF] = 0.01833-s

[H⁺] = s

[F^-] = 0.07638+s

As we know, Ka = [H⁺] [F^‑]/[HF]

7.2X10^-4 = (s) (0.07638+s) / (0.01833-s)

Making Quadratic Equation and solving for the roots of s

s²+ 0.0771s – (1.3976X10^-5) = 0

s₁= 1.7079X10^-4

s₂= -0.0772 (Rejected due to negative Value)

s = 1.7079 X 10^-4 M = [H⁺]

pH = -log(H⁺]

      = -log[1.7079 X 10^-4]

      = 3.7875