${{10}^{-6}}$ 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

Answer Verified Verified
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

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.
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
  & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}} \\
 & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}} \\
Multiplying and diving ${{10}^{-7}}$ by 10 in the above equation, we get
  & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}}\times \frac{10}{10} \\
 & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-8}}\times 10 \\
Taking ${{10}^{-8}}$ common in the equation for simplification, we obtain
  & \left[ O{{H}^{-}} \right]={{10}^{-8}}(1+10) \\
 & \left[ O{{H}^{-}} \right]=11\times {{10}^{-8}} \\
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
  & pOH=-\log \left[ O{{H}^{-}} \right] \\
 & pOH=-\log \left[ 11\times {{10}^{-8}} \right] \\
Applying $\log (mn)=\log m+\log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as
  & pOH=-(\log 11+\log {{10}^{-8}}) \\
 & pOH=-(\log 11-8\log 10) \\
We know that ${{\log }_{10}}10=1$, on substituting it, the above equation becomes
  & pOH=-(\log 11-8) \\
 & pOH=-1.0414+8 \\
 & pOH=6.9586\approx 6.96 \\
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
  & pH+pOH=14 \\
 & pH=14-pOH \\
 & pH=14-6.96 \\
 & pH=7.04 \\
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.