Question

What will be the pH of $0.1M$ NaA solution, If the equilibrium constant of the reaction of weak acid HA with a strong base is ${10^9}$ ?

Answer 1

Hint - To solve this question first we will study all the terms and definitions and then, as we are having all the required values, we will calculate the pH by using the formula of equilibrium constant.
In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations having no further tendency to change with time, so that there is no change in the properties of the system, such a state is known as equilibrium.

Complete answer:
> The equilibrium constant of a reaction (donated as $K$) provides an insight into the relationship between the reactants and products when the reaction is at equilibrium.
> In short, at equilibrium,
Rate (forward reaction) = Rate (backward reaction). i.e.
$ \Rightarrow \,\,{r_f}\, = \,\,\,{r_b}$
> pH – It can be defined as the scale used to state the acidity or basicity of an aqueous solution. Acidic solutions have lower pH values and Basic/Alkaline solutions have higher values of pH. Pure water, at room temperature, is neutral and has a pH of 7.
> Now let’s come to the solution for this question-
 $HA\, + \,NaOH\, \to NaA\, + \,{H_2}O$
Or we can say,
$HA\, + \,O{H^ - }\, \to \,{A^ - }\, + \,{H_2}O$
${K_{eq.}}\, = \,{10^9}\, = \,\,\dfrac{{\left[ {{A^ - }} \right]\left[ {{H_2}O} \right]}}{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}$
Also, $HA\,\, \Leftrightarrow \,\,{H^ + }\, + \,{A^ - }$
${K_a}\, = \,\,\dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Therefore,
$\dfrac{{{K_{eq.}}}}{{{K_a}}}\, = \,\dfrac{1}{{K\omega }}$
Or, we can say that
${K_a}\, = \,{10^9}\, \times \,{10^{ - 14}}\, = \,{10^{ - 5}}$
Thus, for ${A^ - }\, + \,{H_2}O\,\, \Leftrightarrow \,\,HA\,\, + \,\,OH$
$\left[ {O{H^ - }} \right]\, = \,ch$
$ = \,\,C\sqrt {\dfrac{{{K_H}}}{C}} \, = \,\sqrt {\dfrac{{{K_\omega }C}}{{{K_a}}}} $
$ = \,\,\sqrt {\dfrac{{{{10}^{ - 14}}\, \times \,0.1}}{{{{10}^5}}}} \, = \,{10^{ - 5}}$
Therefore, $\left[ {{H^ + }} \right]\,\,\, = \,\,\,{10^9}$
And pH = 9
Hence, the answer is pH = 9

Note – Equilibrium constant can be defined as the ratio of the equilibrium concentrations of products over the equilibrium concentration of reactants each raised to the power of their stoichiometric coefficients. There are different types of equilibrium constants which state the relationship between products and reactants present in equilibrium in terms of Concentration and pressure.