Answer:
Degree of ionization = 0.0377
pH of the solution = 1.769
Explanation:
Initial concentration of HF = 0.45 M
[tex]K_a = 6.8 \times 10^{-4}[/tex]
[tex]HF \leftrightharpoons H^+ + F^-[/tex]
Initial 0.45 0 0
At equi 0.45 - x x x
Equilibrium constant = [tex]\frac{[H^+][F^-]}{HF}[/tex]
[tex]6.8 \times 10^{-4}= \frac{[x][x]}{0.45 - x}[/tex]
[tex]x^2 + 6.8 \times 10^{-4} x - 6.8 \times 10^{-4} \times 4.5 = 0[/tex]
x = 0.017 M
x = Cα
α = Degree of ionization
C = Concentration
Degree of ionization = [tex]\frac{0.017}{0.45} = 0.0377[/tex]
[tex]pH = -log[H^+][/tex]
[H^+]=0.017 M
[tex]pH = -log[0.017][/tex]
= 1.769
