Question

${10^{ - 6}}M$ $NaOH$ is diluted $100$ times. The pH of the diluted base is:
A. between 7 and 8
B. between 5 and 6
C. between 6 and 7
D. between 10 and 11
E. between 9 and 10

Answer 1

Hint: Whenever a solution is diluted, it means its concentration is being decreased accordingly. After diluting the solution 100 times, the concentration of the solution also decreases 100 times. The amount of hydroxyl ions increase on dilution of a solution and the amount of hydrogen ions decrease on dilution.

Complete answer:
Before dilution, the concentration of $NaOH$ solution is = ${10^{ - 6}}M$
After diluting it by $100$ times, the concentration of solution becomes = ${10^{ - 8}}M$
Thus, the concentration of hydroxyl ions becomes = ${10^{ - 8}}M$ .
$\left[ {O{H^ - }} \right] = {10^{ - 8}}M$
Comparatively, the concentration of hydrogen ions will be = ${10^{ - (14 - 8)}} = {10^{ - 6}}M$
Thus, during the dilution process, there is a dissociation of water as well, at room temperature.
${H_2}O \to {H^ + } + O{H^ - }$
This will further increase the concentration of hydroxyl ions in the solution.
Thus, the total concentration of $\left[ {O{H^ - }} \right]$ is equal to:
$\left[ {O{H^ - }} \right] = {10^{ - 7}} + {10^{ - 8}} = 1.1 \times {10^{ - 7}}M$
And the total concentration of $\left[{H^ + } \right]$ is equal to:
$\left[ {{H^ + }} \right] = \dfrac{{{{10}^{ - 14}}}}{{1.1 \times {{10}^{ - 7}}}} = 9.09 \times {10^{ - 8}}M$
Thus, the pH will be = $ - \log (9.09 \times {10^{ - 8}}) = 8 - 0.95 = 7.05$

Thus, the correct option is A. between 7 and 8.

Note:
Whenever a solution is diluted with water, the water itself also contributes some additional hydrogen and hydroxyl ions to the solution which cannot be neglected. Thus, the overall change in pH plays a crucial role in the conductometric and potentiometric titrations conducted in the laboratories.