Question

Calculate the pH of $1.0 \times {10^{ - 3}}M$ sodium phenolate, $NaO{C_6}{H_5}$. ${K_a}$ for $HO{C_6}{H_5}$ is $1.05 \times {10^{ - 10}}$.
A.pH = 3.67
B.pH = 7.78
C.pH = 10.49
D.pH = 12.90

Answer 1

In this question remember to use the concept of Law of mass action which is given as-
${K_c} = \dfrac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}$
here [A] and [b] represents the concentration of reactant and [C] and [D] represents the concentration of product, use this information to approach the solution.

Complete answer:
First let’s write the chemical reaction using the given information so the reaction will be
${C_6}{H_5}{O^ - } + {H_2}O \rightleftharpoons {C_6}{H_5}OH + O{H^ - }$
Now the initial concentration of reactant will be C

{C_6}{H_5}{O^ - } + {H_2}O \rightleftharpoons {C_6}{H_5}OH + O{H^ - } \\\ {\text{ }}C{\text{ }}0{\text{ }}0 \\\ \end{gathered} $$ So, after time t the final concentration will be $$\begin{gathered} {C_6}{H_5}{O^ - } + {H_2}O \rightleftharpoons {C_6}{H_5}OH + O{H^ - } \\\ C\left( {1 - h} \right){\text{ }}Ch{\text{ }}Ch \\\ \end{gathered} $$ Since the reaction is in the state of chemical equilibrium so, by law of mass action ${K_h} = \dfrac{{\left[ {{C_6}{H_5}OH} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{C_6}{H_5}{O^ - }} \right]}}$ ${K_h} = \dfrac{{\left[ {Ch} \right]\left[ {Ch} \right]}}{{\left[ {C\left( {1 - h} \right)} \right]}}$ Now let suppose that h << 1 So, (1 – h) will can be considered as negligible value Therefore, ${K_h} = \dfrac{{\left[ {Ch} \right]\left[ {Ch} \right]}}{{\left[ C \right]}} = \dfrac{{{C^2}{h^2}}}{C}$ ${K_h} = C{h^2}$ Now since $NaO{C_6}{H_5}$ is the salt of weak acid and strong base and we know that this is given as; ${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$ and here ${K_w}$ is constant which is equal to ${10^{ - 14}}$ and since we have ${K_a}$ = $1.05 \times {10^{ - 10}}$ Now substituting the values in the formula ${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$ we get ${K_h} = \dfrac{{{{10}^{ - 14}}}}{{1.05 \times {{10}^{ - 10}}}} = 9.5 \times {10^{ - 5}}$ or $C{h^2} = 9.5 \times {10^{ - 5}}$ Now we know that value of C = $1.0 \times {10^{ - 3}}M$ So, we got $1.0 \times {10^{ - 3}} \times {h^2} = 9.5 \times {10^{ - 5}}$ ${h^2} = 9.5 \times {10^{ - 2}}$ $h = \sqrt {9.5 \times {{10}^{ - 2}}} = 3.08 \times {10^{ - 1}}$ Now we know that $\left[ {O{H^ - }} \right] = Ch$ So, substituting the value in the above equation we get $\left[ {O{H^ - }} \right] = 1.0 \times {10^{ - 3}} \times 3.08 \times {10^{ - 1}} = 3.08 \times {10^{ - 4}}M$ Now we know that $pOH = - \log \left( {\left[ {O{H^ - }} \right]} \right)$ Therefore, $pH = - \log \left( {3.08 \times {{10}^{ - 4}}} \right) = 3.51$ Now we know that pH + pOH = 14 Therefore, pH = 14 – pOH pH = 14 – 3.51 = 10.49 so, the pH of sodium phenolate is 10.49 . **Hence, option C is the correct option.** **Note:** In the above solution we came across the term “Law of mass action” which is a concept which explains the behavior of solution at the condition of dynamic equilibria. According to this law rate of a reaction is directly proportional to the product of concentrations of reactants in the given reaction.