Question

The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.

Answer 1

Approximately
`8.3 * 10^(-2)\; \rm mol \cdot L^(-1)`.

Step-by-step explanation:

The
`K_a` in this question refers the dissociation equilibrium of
`\rm HC_6H_4ClO` as an acid:

`\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^(+) \, (aq) + C_6H_4ClO^(-)\, (aq)`.

`\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]}`.

However, the question also states that the solution here has a
`\rm pH` of
`11.05`, which means that this solution is basic. In basic solutions at
`\rm 25\;^\circ C`, the concentration of
`\rm H^(+)` ions is considerably small (typically less than
`10^(-7)\;\rm mol \cdot L^(-1)`.) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of
`\rm H^(+)` ions.

Here's the workaround: note that
`\rm C_6H_4ClO^(-)\, (aq)` is the conjugate base of the weak acid
`\rm HC_6H_4ClO\, (aq)`. Therefore, when
`\rm C_6H_4ClO^(-)\, (aq)` dissociates in water as a base, its
`K_b` would be equal to
`\displaystyle (K_w)/(K_a) \approx (10^(-14))/(K_a)`. (
`K_w` is the self-ionization constant of water.
`K_w \approx 10^(-14)` at
`\rm 25\;^\circ C`.)

In other words,

`\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^(-)}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx (10^(-14))/(6.6 * 10^(-10)) \\ & \approx 1.51515 * 10^(-5)\end{aligned}`.

And that
`K_b` value corresponds to the equilibrium:

`\rm C_6H_4ClO^(-)\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^(-)\, (aq)`.

`\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]}`.

The value of
`K_b` has already been found.

The
`\rm OH^(-)` concentration of this solution can be found from its
`\rm pH` value:

`\begin{aligned}& \left[\mathrm{OH^(-)}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^(+)\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx (10^(-14))/(10^(-11.05)) \\ &\approx 1.1220 * 10^(-3)\; \rm mol\cdot L^(-1) \end{aligned}`.

To determine the concentration of
`\left[\mathrm{HC_6H_4ClO}\right]`, consider the following table:

`\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}`

Before hydrolysis, the concentration of both
`\mathrm{HC_6H_4ClO}` and
`\rm OH^(-)` are approximately zero. Refer to the chemical equation. The coefficient of
`\mathrm{HC_6H_4ClO}` and
`\mathrm{HC_6H_4ClO}` are the same. As a result, this equilibrium will produce
`\rm OH^(-)` and
`\mathrm{HC_6H_4ClO}` at the exact same rate. Therefore, at equilibrium,
`\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^(-)}\right] \approx 1.1220 * 10^(-3)\; \rm mol\cdot L^(-1)`.

Calculate the equilibrium concentration of
`\left[\mathrm{C_6H_4ClO^(-)}\right]` from
`K_b\left(\mathrm{C_6H_4ClO^(-)}\right)`:

`\begin{aligned} & \left[\mathrm{C_6H_4ClO^(-)}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^(-)}\right]}{K_b}\\&\approx (\left(1.1220 * 10^(-3)\right) * \left(1.1220 * 10^(-3)\right))/(1.51515* 10^(-5))\; \rm mol \cdot L^(-1) \\ &\approx 8.3 * 10^(-2)\; \rm mol \cdot L^(-1)\end{aligned}`.