Answer:
[tex]\displaystyle K_a = 4.24\times 10^{-10}[/tex]
Explanation:
Write the base reaction of NaA with water:
[tex]\displaystyle \text{A}^-_\text{(aq)}+\text{H$_2$O}_\text{($\ell$)}\rightleftharpoons \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)}[/tex]
Hence, the equilibrium constant expression for the reaction is:
[tex]\displaystyle K_b = \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]}[/tex]
Thus, to find Ka, we can find Kb and use the fact that Ka × Kb = Kw.
From the reaction and initial concentration of NaA, create an ICE chart:
[tex]\begin{tabular}{llllll} & A^- &\text{H$_2$O} & \rightleftharpoons & HA & OH^- \\I & 0.563 M & \---- & & 0 M & 0 M \\C & -\text{ $ x$} & \---- & & +\text{ $x$ M} & + \text{$x$ M} \\E & \text{(0.563 - $x$) M} & \---- & & \text{$x$ M} & \text{$x$ M} \end{tabular}[/tex]
Find [OH⁻] from the given pH:
[tex]\displaystyle \begin{aligned} \text{pH} +\text{pOH} & = 14.00 \\ \\ \text{pOH} & = 14.00 - \text{pH} \\ \\ & = 14.00 - (11.56) \\ \\ & = 2.44 \\ \\ -\log[\text{OH}^-] & = 2.44 \\ \\ [\text{OH}^-] &= 10^{-2.44} \\ \\ & =0.00363 \text{ M}= 3.63\times 10^{-3} \text{ M} = x\text{ M}\end{aligned}[/tex]
Solve for all species concentrations at equilibrium from the found x value:
[tex]\displaystyle [\text{HA}] = [\text{OH}^-] = 3.63\times 10^{-3} \text{ M}[/tex]
And:
[tex]\displaystyle \begin{aligned} \ [\text{A}^-] & = 0.563 - 3.63\times 10^{-3} \text{ M}\\ \\ & = 0.559\text{ M}\end{aligned}[/tex]
Find Kb:
[tex]\displaystyle \begin{aligned} \displaystyle K_b &= \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]} \\ \\ & = \frac{(3.63\times 10^{-3})(3.63\times 10^{-3})}{(0.559)}\\ \\ & = 2.36\times 10^{-5}\end{aligned}[/tex]
Find Ka:
[tex]\displaystyle \begin{aligned} K_a\cdot K_b & = K_w \\ \\ K_a & = \frac{K_w}{K_b} \\ \\ & = \frac{(1.00 \times 10^{-14})}{(2.36\times 10^{-5})} \\ \\ &= 4.24\times 10^{-10} \end{aligned}[/tex]
In conclusion:
[tex]\displaystyle K_a = 4.24\times 10^{-10}[/tex]
