Answer
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Hint:
Think about how the $pH$ of any solution depends on the hydrogen and hydroxide ions present in the solution on dilution. Consider the relation between concentration of hydrogen ions and $pH$ value.
Complete step by step solution:
First of all, we have the pH of the initial acid solution is 6. Consider the formula relating the concentration of hydrogen ions and the $pH$.
\[pH=-\log [{{H}^{+}}]\]
Thus, to find the initial concentration of the hydrogen ions:
\[6=-\log [{{H}^{+}}]\]
Now, solving for $[{{H}^{+}}]$, we get:
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=1}{{\text{0}}^{\text{-6}}}$ M
After dilution of the acid solution 1000 times,
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\dfrac{\text{1}{{\text{0}}^{\text{-6}}}}{\text{1}{{\text{0}}^{\text{3}}}}\text{=1}{{\text{0}}^{\text{-9}}}\text{M}$
Under this condition of dilution, we cannot neglect the hydrogen ion concentration from water molecules. That is:
${{\text{H}}_{\text{2}}}\text{O}\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+O}{{\text{H}}^{-}}$
Here, we know that, $\left[ {{\text{H}}^{\text{+}}} \right]\text{=1}{{\text{0}}^{\text{-7}}}$
So, total concentration of hydrogen ions is:
Total $\left[ {{H}^{+}} \right]={{10}^{-9}}+{{10}^{-7}}$
We can take ${{10}^{-7}}$ outside as common,
$\begin{align}
& ={{10}^{-7}}\left( 10{}^{-2}+1 \right) \\
& ={{10}^{-7}}\left( \dfrac{101}{100} \right) \\
& ={{10}^{-7}}\times 1.01 \\
\end{align}$
This is the final total concentration of the hydrogen ions, putting this value in the formula to find the $pH$.
$\begin{align}
& pH=-\log \left[ {{H}^{+}} \right]=-\log \left[ 1.01\times {{10}^{-7}} \right] \\
& =-\log (1.01)-\log \left( {{10}^{-7}} \right) \\
& =-0.01+7 \\
& =6.99 \\
\end{align}$
So, the pH of the final solution will be 6.99. Therefore, correct option is ‘D. 6.99’
Additional Information:
- The ionic equilibrium is the equilibrium established in weak electrolytes in between the undissociated ions and dissociated ions. We can see this in the dissociation of water in the question.
- We know that the pH is the measure of concentration of hydrogen ions in a solution. It is also a measure of alkalinity or acidity of a solution.
Note:
When you are attempting these types of problems, it is essential to consider the concentration of hydrogen ions that is contributed by the water molecules used for dilution. It is about ${{10}^{-7}}$M considering the degree of dissociation and $pH$ of water, which is 7. It will increase the acidity of the solution.
Think about how the $pH$ of any solution depends on the hydrogen and hydroxide ions present in the solution on dilution. Consider the relation between concentration of hydrogen ions and $pH$ value.
Complete step by step solution:
First of all, we have the pH of the initial acid solution is 6. Consider the formula relating the concentration of hydrogen ions and the $pH$.
\[pH=-\log [{{H}^{+}}]\]
Thus, to find the initial concentration of the hydrogen ions:
\[6=-\log [{{H}^{+}}]\]
Now, solving for $[{{H}^{+}}]$, we get:
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=1}{{\text{0}}^{\text{-6}}}$ M
After dilution of the acid solution 1000 times,
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\dfrac{\text{1}{{\text{0}}^{\text{-6}}}}{\text{1}{{\text{0}}^{\text{3}}}}\text{=1}{{\text{0}}^{\text{-9}}}\text{M}$
Under this condition of dilution, we cannot neglect the hydrogen ion concentration from water molecules. That is:
${{\text{H}}_{\text{2}}}\text{O}\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+O}{{\text{H}}^{-}}$
Here, we know that, $\left[ {{\text{H}}^{\text{+}}} \right]\text{=1}{{\text{0}}^{\text{-7}}}$
So, total concentration of hydrogen ions is:
Total $\left[ {{H}^{+}} \right]={{10}^{-9}}+{{10}^{-7}}$
We can take ${{10}^{-7}}$ outside as common,
$\begin{align}
& ={{10}^{-7}}\left( 10{}^{-2}+1 \right) \\
& ={{10}^{-7}}\left( \dfrac{101}{100} \right) \\
& ={{10}^{-7}}\times 1.01 \\
\end{align}$
This is the final total concentration of the hydrogen ions, putting this value in the formula to find the $pH$.
$\begin{align}
& pH=-\log \left[ {{H}^{+}} \right]=-\log \left[ 1.01\times {{10}^{-7}} \right] \\
& =-\log (1.01)-\log \left( {{10}^{-7}} \right) \\
& =-0.01+7 \\
& =6.99 \\
\end{align}$
So, the pH of the final solution will be 6.99. Therefore, correct option is ‘D. 6.99’
Additional Information:
- The ionic equilibrium is the equilibrium established in weak electrolytes in between the undissociated ions and dissociated ions. We can see this in the dissociation of water in the question.
- We know that the pH is the measure of concentration of hydrogen ions in a solution. It is also a measure of alkalinity or acidity of a solution.
Note:
When you are attempting these types of problems, it is essential to consider the concentration of hydrogen ions that is contributed by the water molecules used for dilution. It is about ${{10}^{-7}}$M considering the degree of dissociation and $pH$ of water, which is 7. It will increase the acidity of the solution.
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