Question

A solution contains 0.45 M hydrofluoric acid (HF; KA = 6.8 X 10−4). Write the dissociation reaction. Determine the degree of ionization and the pH of the solution

Answer 1

Answer:

Degree of ionization = 0.0377

pH of the solution = 1.769

Explanation:

Initial concentration of HF = 0.45 M

$K_a = 6.8 \times 10^{-4}$

                     $HF \leftrightharpoons H^+ + F^-$

Initial        0.45                             0          0

At equi      0.45 - x                      x           x

Equilibrium constant = $\frac{[H^+][F^-]}{HF}$

                   $6.8 \times 10^{-4}= \frac{[x][x]}{0.45 - x}$

           $x^2 + 6.8 \times 10^{-4} x - 6.8 \times 10^{-4} \times 4.5 = 0$

x = 0.017 M

x = Cα

α = Degree of ionization

C = Concentration

Degree of ionization = $\frac{0.017}{0.45} = 0.0377$

$pH = -log[H^+]$

[H^+]=0.017 M

$pH = -log[0.017]$

             = 1.769