Question

What volume of $NaOH$ solution having pH=11 should be added in 1 litre of 0.1 M $HCl$ solution to increase its pH by 2 units?
(A) 54 litres
(B) 49.5 litre
(C) 62.7 litres
(D) 98 litres

Answer 1

Hint: The pH scale deals with how alkaline or acidic a substance is and the pH scale runs from 0 to 14, with a pH higher than seven being alkaline, a pH lower than seven being acidic, and a pH of seven being neutral. Also, each whole pH value below seven will be ten times more acidic than the next higher value which means that the pH scale is logarithmic.

Complete step by step solution:
- In the question, to a one litre solution whose molarity is 0.1 M, $NaOH$ solution with a pH value 11 is being added. We are asked to find the volume of sodium hydroxide needed to increase the pH of the solution by two units. Let’s answer this question by step by step.
- The pH of the initial solution (0.1 M 1 litre $HCl$) can be determined as follows
\[Initial~pH\text{ }=-~\text{log}\left[ {{H}^{+}} \right]=-~\text{log}\left[ {{10}^{-1}} \right]\]
\[~pH\text{ }=1\]
As in the question the pH should increase to 3.Then the $HCl$ concentration should be 0.001 and hence we can write the concentration of hydroxyl ion in $NaOH$ as follows
\[{{10}^{-3}}=\left[ O{{H}^{-}} \right]\text{ }\left( \because pOH=3 \right)\]
 Now let’s find the millimoles of ${{H}^{+}}$ and $NaOH$ as follows
\[millimoles\text{ }of\text{ }{{H}^{+}}=0.1\times 1000\text{ }ml=100\]
\[millimoles\text{ }of\text{ }NaOH=V\times 0.001\]
Where V is the volume of $NaOH$ needed to add to $HCl$. The final concentration of $HCl$ can be given by the following relation
\[\dfrac{100-V\times 0.001}{V+1000}=0.001\]
Rearrange the above equation as follows,
\[100=2V\times 0.001+1\]
\[2V=\dfrac{99}{0.001}\]
\[V=49.5\times {{10}^{3}}mL\]
Converting mL to litre we get,
\[V=49.5L\]

Therefore the correct option is (B) 49.5 litre.

Note: It should be noted that the pH of an acidic solution is lesser than 7.As we add $NaOH$ to acidic $HCl$, the pH moves closer to the pH of pure water, and pH of alkalis (above pH 7). Thus the pH of an acidic solution such as $HCl$ will increase in addition to $NaOH$ and the acidity will decrease.