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**Hint:**pH of a solution is calculated from the concentration of ${{H}^{+}}$ ions in the solution and is given as $pH=-\log \left[ {{H}^{+}} \right]$.

Similarly, pOH measures $O{{H}^{-}}$ ion concentration in a solution, i.e. $pOH=-\log \left[ O{{H}^{-}} \right]$.

Relationship between pH and pOH, based on the equilibrium concentrations of ${{H}^{+}}$ and $O{{H}^{-}}$ is

\[pH+pOH=14\]

**Complete answer:**

Let us solve the given question step by step.

Given, molar concentration of NaOH base = ${{10}^{-6}}$ M

NaOH being a strong base dissociates completely in water into $N{{a}^{+}}$ and $O{{H}^{-}}$ ions

$NaOH(aq)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)$

Since, one mole of NaOH is equal to one of $N{{a}^{+}}$ and $O{{H}^{-}}$. Thus, we can draw the following conclusion

\[\left[ NaOH \right]=\left[ N{{a}^{+}} \right]=\left[ O{{H}^{-}} \right]={{10}^{-6}}M\]

It is given that the base has been diluted by 100 times. So, the concentration of $\left[ O{{H}^{-}} \right]$now becomes ${{10}^{-8}}M$, i.e.

\[\dfrac{{{10}^{-6}}M}{100}={{10}^{-8}}M\]

To find the total concentration of $O{{H}^{-}}$ ions, i.e. $\left[ O{{H}^{-}} \right]={{10}^{-8}}M+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$, we need to consider the concentration of $O{{H}^{-}}$ from water.

We know that water ionizes as

\[{{H}_{2}}O\rightleftarrows {{H}^{+}}+O{{H}^{-}}\]

One mole of water gives one mole of ${{H}^{+}}$ and $O{{H}^{-}}$,thus, we have

\[\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]\]

Ionic product of water, which is the product of concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ ions, at 25$^{o}C$ is ${{10}^{-14}}$. Since \[\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}\], we can write that the concentration of $O{{H}^{-}}$ from ionization of water, ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}={{10}^{-7}}M$.

Therefore, total $\left[ O{{H}^{-}} \right]$ ions after substituting the value of ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$ will be

$\begin{align}

& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}} \\

& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}} \\

\end{align}$

Multiplying and diving ${{10}^{-7}}$ by 10 in the above equation, we get

$\begin{align}

& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}}\times \frac{10}{10} \\

& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-8}}\times 10 \\

\end{align}$

Taking ${{10}^{-8}}$ common in the equation for simplification, we obtain

$\begin{align}

& \left[ O{{H}^{-}} \right]={{10}^{-8}}(1+10) \\

& \left[ O{{H}^{-}} \right]=11\times {{10}^{-8}} \\

\end{align}$

Now we have the total concentration of $O{{H}^{-}}$, i.e. $\left[ O{{H}^{-}} \right]=11\times {{10}^{-8}}M$, we can find pOH as

$\begin{align}

& pOH=-\log \left[ O{{H}^{-}} \right] \\

& pOH=-\log \left[ 11\times {{10}^{-8}} \right] \\

\end{align}$

Applying $\log (mn)=\log m+\log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as

\[\begin{align}

& pOH=-(\log 11+\log {{10}^{-8}}) \\

& pOH=-(\log 11-8\log 10) \\

\end{align}\]

We know that ${{\log }_{10}}10=1$, on substituting it, the above equation becomes

\[\begin{align}

& pOH=-(\log 11-8) \\

& pOH=-1.0414+8 \\

& pOH=6.9586\approx 6.96 \\

\end{align}\]

To find the pH from pOH, we have the relation that is true for solution at 25$^{o}C$. Putting the value of pOH = 6.96, we have the pH of the solution

\[\begin{align}

& pH+pOH=14 \\

& pH=14-pOH \\

& pH=14-6.96 \\

& pH=7.04 \\

\end{align}\]

Therefore, the pH of diluted base is 7.04, which lies within 7 to 8.

**So, the correct answer is “Option C”.**

**Note:**Note that concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ is important for dilute solutions. We cannot ignore the concentration of $O{{H}^{-}}$ due to water in this case, as the solution has become very dilute due to the addition of water. Due to the decrease in the number of $O{{H}^{-}}$ (and ${{H}^{+}}$) ions per unit volume, the pH of the basic solution has been reduced.

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