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${{10}^{-6}}$ M NaOH is diluted by 100 times. The pH of diluted base is?(A)- Between 6 & 7(B)- Between 10 & 11(C)- Between 7 & 8(D)- Between 5 & 6

Last updated date: 10th Sep 2024
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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
$pH+pOH=14$

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.