Question

The hydrogen ion concentration of a ${{10}^{-8}}$ $HCl$ aqueous solution at $298K$ is:
[Given that: ${{K}_{W}}={{10}^{-14}}$]
A. $1.0\times {{10}^{-6}}M$
B. $1.0525\times {{10}^{-7}}M$
C. $9.525\times {{10}^{-8}}M$
D. $1.0\times {{10}^{-8}}M$

Answer 1

Hint: The total concentration of hydrogen can be found out by adding the hydrogen ion in the dissociation of acid, and the hydrogen ion that is present in the water as ${{H}_{3}}{{O}^{+}}$.

Complete step by step solution:
In order to answer our question, we need to learn about the ionisation of acids and bases. According to Arrhenius concept they are strong acids and bases as they are able to completely dissociate and produce ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ ions respectively in the medium. Alternatively, the strength of an acid or base may also be gauged in terms of Bronsted-Lowry concept of acids and bases, where in a strong base implies a good proton acceptor and a strong acid means a good proton donor. Let us see the acid-base dissociation equilibrium of a weak acid, HA.
\[HA+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+{{A}^{-}}\]
The acid or base dissociation equilibrium is dynamic involving a transfer of proton in forward and reverse directions. Now, the question arises that in case the equilibrium is dynamic then as time passes, which direction is favoured? What is the driving force behind it? In order to answer these questions we shall deal into the issue of comparing the strengths of the two acids or bases involved in the dissociation equilibrium. Water is a weak electrolyte, it is dissociated as:
\[{{H}_{2}}O\rightleftharpoons {{H}^{+}}+O{{H}^{-}}\]
So, applying law of equilibrium, we have $K=\dfrac{{{[H]}^{+}}{{[OH]}^{-}}}{[{{H}_{2}}O]}$, as $[{{H}_{2}}O]$ is constant, so we can write ${{K}_{W}}={{[H]}^{+}}{{[OH]}^{-}}$.
Now, talking about hydrogen ion concentration, they are present both in the acid after it gets dissociated and also in water dissociation. Now, HCl dissociates as $HCl\rightleftharpoons {{H}^{+}}+C{{l}^{-}}$, as all are 1 molar to concentration of ${{H}^{+}}$ is ${{10}^{-8}}$
For finding concentration from water, we will use the equation for ${{K}_{W}}$, so:${{[H]}^{+}}{{[OH]}^{-}}={{10}^{-14}}\Rightarrow {{[H]}^{+}}={{10}^{-14-(-8)}}={{10}^{-7}}M$
So, the total concentration is $({{10}^{-8}}+{{10}^{-7}})M=11\times {{10}^{-8}}M=1.10\times {{10}^{-7}}M$.

So we get our answer approximately close to option (B).

Note: Acidic, neutral or basic aqueous solutions can be distinguished by the relative values of ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ concentrations.
\[\begin{align}
& Acidic\,\,\,\,\,\,\,{{[{{H}_{3}}O]}^{+}}{{[OH]}^{-}} \\
& Neutral\,\,\,\,{{[{{H}_{3}}O]}^{+}}{{[OH]}^{-}} \\
& Basic\,\,\,\,\,\,\,\,\,{{[{{H}_{3}}O]}^{+}}{{[OH]}^{-}} \\
\end{align}\]