Respuesta :
Answer:
Approximately [tex]8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}[/tex].
Explanation:
The [tex]K_a[/tex] in this question refers the dissociation equilibrium of [tex]\rm HC_6H_4ClO[/tex] as an acid:
[tex]\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)[/tex].
[tex]\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}[/tex].
However, the question also states that the solution here has a [tex]\rm pH[/tex] of [tex]11.05[/tex], which means that this solution is basic. In basic solutions at [tex]\rm 25\;^\circ C[/tex], the concentration of [tex]\rm H^{+}[/tex] ions is considerably small (typically less than [tex]10^{-7}\;\rm mol \cdot L^{-1}[/tex].) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of [tex]\rm H^{+}[/tex] ions.
Here's the workaround: note that [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] is the conjugate base of the weak acid [tex]\rm HC_6H_4ClO\, (aq)[/tex]. Therefore, when [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] dissociates in water as a base, its [tex]K_b[/tex] would be equal to [tex]\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}[/tex]. ([tex]K_w[/tex] is the self-ionization constant of water. [tex]K_w \approx 10^{-14}[/tex] at [tex]\rm 25\;^\circ C[/tex].)
In other words,
[tex]\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}[/tex].
And that [tex]K_b[/tex] value corresponds to the equilibrium:
[tex]\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)[/tex].
[tex]\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}[/tex].
The value of [tex]K_b[/tex] has already been found. Â
The [tex]\rm OH^{-}[/tex] concentration of this solution can be found from its [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}[/tex].
To determine the concentration of [tex]\left[\mathrm{HC_6H_4ClO}\right][/tex], consider the following table:
[tex]\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}[/tex]
Before hydrolysis, the concentration of both [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\rm OH^{-}[/tex] are approximately zero. Refer to the chemical equation. The coefficient of [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] are the same. As a result, this equilibrium will produce [tex]\rm OH^{-}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] at the exact same rate. Therefore, at equilibrium, [tex]\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}[/tex].
Calculate the equilibrium concentration of [tex]\left[\mathrm{C_6H_4ClO^{-}}\right][/tex] from [tex]K_b\left(\mathrm{C_6H_4ClO^{-}}\right)[/tex]:
[tex]\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
