Question

${10}^{-6}$ 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 1

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.
     $${\displaystyle \frac{{10}^{-6}M}{100}}={10}^{-8}M$$
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
     $\begin{array}{rl}& \left[O{H}^{-}\right]={10}^{-8}+{\left[O{H}^{-}\right]}_{H_{2}O}\\ & \left[O{H}^{-}\right]={10}^{-8}+{10}^{-7}\end{array}$
Multiplying and diving ${10}^{-7}$ by 10 in the above equation, we get
     $\begin{array}{rl}& \left[O{H}^{-}\right]={10}^{-8}+{10}^{-7}\times \frac{10}{10}\\ & \left[O{H}^{-}\right]={10}^{-8}+{10}^{-8}\times 10\end{array}$
Taking ${10}^{-8}$ common in the equation for simplification, we obtain
     $\begin{array}{rl}& \left[O{H}^{-}\right]={10}^{-8}(1+10)\\ & \left[O{H}^{-}\right]=11\times {10}^{-8}\end{array}$
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
     $\begin{array}{rl}& pOH=-\log \left[O{H}^{-}\right]\\ & pOH=-\log [11\times {10}^{-8}]\end{array}$
Applying $\log (mn)=\log m+\log n$ and $\log m^n=n\log m$, we can simplify the above equation as
     $$\begin{array}{rl}& pOH=-(\log 11+\log {10}^{-8})\\ & pOH=-(\log 11-8\log 10)\end{array}$$
We know that ${\log }_{10}10=1$, on substituting it, the above equation becomes
     $$\begin{array}{rl}& pOH=-(\log 11-8)\\ & pOH=-1.0414+8\\ & pOH=6.9586\approx 6.96\end{array}$$
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
     $$\begin{array}{rl}& pH+pOH=14\\ & pH=14-pOH\\ & pH=14-6.96\\ & pH=7.04\end{array}$$
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.