Answers: 1
.
in this question refers the dissociation equilibrium of 
as an acid:
. ![\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]}](/tpl/images/0619/7247/8ffd8.png)
.
of 
, which means that this solution is basic. In basic solutions at 
, the concentration of 
ions is considerably small (typically less than 
.) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of 
is the conjugate base of the weak acid 
. Therefore, when 
would be equal to 
. (
is the self-ionization constant of water. 
at 
.
.![\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]}](/tpl/images/0619/7247/982ba.png)
.
concentration of this solution can be found from its ![\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}](/tpl/images/0619/7247/8b5c5.png)
.![\left[\mathrm{HC_6H_4ClO}\right]](/tpl/images/0619/7247/a50f4.png)
, consider the following table:
and ![\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}](/tpl/images/0619/7247/7c60d.png)
.![\left[\mathrm{C_6H_4ClO^{-}}\right]](/tpl/images/0619/7247/1ccf2.png)
from 
:![\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}](/tpl/images/0619/7247/02e65.png)
.
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The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05...
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