Answer:
Ka = [tex]2.32 \times 10^{-4}[/tex]
pH = 2.12
Explanation:
Calculation of Ka:
% Dissociation = 3% = 0.03
Concentration of solution = 0.25 M
HA dissociated as:
[tex]HA \rightarrow H^+ + A^{-}[/tex]
C(1 - 0.03) C×0.03 C×0.03
HA] after dissociation = 0.25×0.97 = 0.2425 M
[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]
[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]
[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]
[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]
Calculation of pH of the solution
[tex]pH = -log [H^+][/tex]
[tex]H^+[\tex] = 0.0075 M[/tex]
[tex]pH = -log 0.0075 = 2.12[/tex]
