Answer:
Ka = [tex]4.04 \times 10^{-11}[/tex]
Explanation:
Initial concentration of weak acid = [tex]4.5 \times 10^{-4}\ M[/tex]
pH = 6.87
[tex]pH = -log[H^+][/tex]
[tex][H^+]=10^{-pH}[/tex]
[tex][H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M[/tex]
HA dissociated as:
[tex]HA \leftrightharpoons H^+ + A^{-}[/tex]
(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = [tex]1.35 \times 10^{-7}\ M[/tex]
[tex]Ka = \frac{[H^+][A^{-}]}{[HA]}[/tex]
[tex]Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}[/tex]
0.000000135 <<< 0.00045
[tex]Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}[/tex]
