Question

What is the pH of a solution in which 25.0 mL of a 0.100 M solution of NaOH has been added to 100 mL of a 0.100 M HCl solution?

Answer 1

Hint: The pH of a solution is the measure of the concentration of hydronium ions present in the solution. It is the minus logarithm of the concentration of hydrogen ion:
\[\text{pH}=-\log \left[ {{\text{H}}^{+}} \right]\]
In an acid-base reaction, hydrogen ion from acid neutralizes hydroxyl ion from the base and forms salt and water. The excess of either acid or base results in the change in pH of the solution.

Complete answer:
The reaction involved in this process is written below.
\[\text{HCl}+\text{NaOH}\to \text{NaCl}+{{\text{H}}_{\text{2}}}\text{O}\]
According to the balanced chemical reaction, sodium hydroxide and hydrochloric acid neutralize each other in $1:1$ ratio. The molarity of the two solutions is the same but their volume is different. We can find the number of moles of each by using the given formula:
\[\text{Molarity (M)}=\dfrac{\text{Number of moles (n)}}{\text{Volume of solution in litre (V)}}\]
Since HCl is a strong acid, we can assume that it completely dissociates. The initial concentration of hydrogen ion $\left( {{\text{H}}^{+}} \right)$ from HCl is 0.100 M. The $\left( \text{O}{{\text{H}}^{-}} \right)$ from base completely neutralizes the solution, and only the excess of one of those will determine the pH of the solution.
Thus, the solution was initially acidic before the addition of base, and the initial amount of hydrogen ion is:
\[\begin{align}
  & 0.100\text{ M}=\dfrac{\text{n}_{{{\text{H}}^{+}}}^{\text{o}}\times 1000}{100} \\
 & \Rightarrow \text{n}_{{{\text{H}}^{+}}}^{\text{o}}=\dfrac{0.100}{10}=0.0100\text{ moles} \\
\end{align}\]
Once 25 mL of 0.100 M NaOH is added, the number of moles of hydroxyl ions produced is:
\[\begin{align}
  & 0.100\text{M}=\dfrac{\text{n}_{\text{O}{{\text{H}}^{-}}}^{\text{o}}\times 1000}{25} \\
 & \Rightarrow \text{n}_{\text{O}{{\text{H}}^{-}}}^{\text{o}}=\dfrac{0.100}{40}=0.0025\text{ moles} \\
\end{align}\]
The total volume of solution is: $100\text{ mL}+25\text{ mL}=125\text{ mL}=0.125\text{ L}$
The number of moles of hydrogen ions left after completely neutralizing NaOH:
\[\begin{align}
  & {{\text{n}}_{{{\text{H}}^{+}}}}=\text{n}_{{{\text{H}}^{+}}}^{\text{o}}-\text{n}_{_{\text{NaOH}}}^{\text{o}} \\
 & \Rightarrow {{\text{n}}_{{{\text{H}}^{+}}}}=\text{0}\text{.0100}-\text{0}\text{.0025} \\
 & \Rightarrow {{\text{n}}_{{{\text{H}}^{+}}}}=\text{0}\text{.0075 moles} \\
\end{align}\]
So, the concentration of ${{\text{H}}^{+}}$ ion is:
\[\begin{align}
  & \left[ {{\text{H}}^{+}} \right]=\dfrac{\text{Number of moles }\left( {{\text{n}}_{{{\text{H}}^{+}}}} \right)}{\text{Volume of solution}} \\
 & \Rightarrow \left[ {{\text{H}}^{+}} \right]=\dfrac{0.0075\text{ moles}}{\text{0}\text{.125 L}}=0.06\text{ M} \\
\end{align}\]
Now, pH is calculated as follows:
\[\begin{align}
  & \text{pH}=-\log \left[ {{\text{H}}^{+}} \right] \\
 & \Rightarrow \text{pH}=-\log \left( 0.06 \right) \\
 & \Rightarrow \text{pH}=-(-1.22) \\
 & \Rightarrow \text{pH}=1.22 \\
\end{align}\]
Hence, the pH of the solution is 1.22 which means that solution is highly acidic.

Note:
In this reaction, NaOH is acting as a limiting reagent which means that it is completely used up when the reaction is completed. And the remaining amount of HCl which did not react with NaOH determines the pH of the solution.