Question

A certain weak acid has ${K_a} = {10^{ - 5}}$. If the equilibrium constant for its reaction with a strong base is represented by $y \times {10^{10}}$, then find the value of y.

Answer 1

We know that, equilibrium constant $\left( {{K_{eq}}} \right)$ is a constant that indicates the side (reactant or product) of an equilibrium favored under certain conditions. The larger the value of K means equilibrium favored in the product side and if K is a smaller number, equilibrium is favored in the reactant side.

Complete step by step answer: Given that a certain weak acid has ${K_a} = {10^{ - 5}}$. Let’s take the weak acid as HX. It undergoes dissociation to form hydrogen ions. The reaction can be written as follows:

$HX\left ( aq \right )\rightleftharpoons H^{+}\left ( aq \right )+X^{-} K_{a}=10^{-5}$ …… (1)

Next, given that, this weak acid reacts with a strong base. Then the equilibrium concentration for reaction is $y \times {10^{10}}$. We know that a strong base is the compound that can release hydroxide ion in aqueous solution. So, the reaction of acid with base can be shown as below.

$H^{+}\left ( aq \right )+OH^{-}\left ( aq \right )\rightleftharpoons H_{2}O\left ( l \right )$ ……(2)

The above reaction is the reverse of autoprotolysis of water, that is,

$H_{2}O\left ( l \right )\rightleftharpoons H^{+}\left ( aq \right )+OH^{-}\left ( aq \right ) …… K=10^{-14}$

So, for equation (2), value of K$= \dfrac{1}{{{{10}^{ - 14}}}} = {10^{14}}$

$H^{+}\left ( aq \right )+OH^{-}\left ( aq \right )\rightleftharpoons H_{2}O\left ( l \right ) K=10^{14}$ …… (3)

Now, we have to add equation (1) and (3).

$HX\left ( aq \right )+OH^{-}\left ( aq \right )\rightleftharpoons H_{2}O\left ( l \right )+X^{-} K_{eq}$

To calculate equilibrium constant, we have to multiply the K value of both the reactions.

$\Rightarrow {K_{eq}} = {10^{ - 5}} \times {10^{14}} = {10^9}$

Given that, equilibrium constant of the reaction is $y \times {10^{10}}$.

So,

$\Rightarrow y \times {10^{10}} = 1 \times {10^9}$

$\Rightarrow y = 0.1$

Hence, the value of y is 0.1.

Additional information:
Let’s understand autoprotolysis of water. When water molecules dissociate to produce hydrogen ion and hydroxide ion, this reaction is called autoprotolysis of water. This reaction is also termed as self ionization water.
$H_{2}O\left ( l \right )\rightleftharpoons H^{+}\left ( aq \right )+OH^{-}\left ( aq \right )$

Note: Always remember that the equilibrium constant of a reverse reaction is reciprocal of the equilibrium constant for the corresponding forward reaction. For the reaction,
$H_{2}O\left ( l \right )\rightleftharpoons H^{+}\left ( aq \right )+OH^{-}\left ( aq \right )$
Equilibrium constant is ${10^{ - 14}}$
For the reverse reaction,
$H^{+}\left ( aq \right )+OH^{-}\left ( aq \right )\rightleftharpoons H_{2}O\left ( l \right )$
Equilibrium constant is $= \dfrac{1}{{{{10}^{ - 14}}}} = {10^{14}}$