Question

What is the pH of a solution made by mixing 100.0 mL of 0.10 M $\text{HN}{{\text{O}}_{3}}$, 50.0 mL of 0.20 M $\text{HCl}$ and 100.0 mL of water? Assume that the volumes are additive.

Answer 1

pH is a measure of the acidity or alkalinity of a solution. It stands for the negative logarithmic function of hydronium ion or hydrogen ion concentration:
$\text{pH}=-\log \left[ {{\text{H}}^{+}} \right]\text{ or }-\log \left[ {{\text{H}}_{\text{3}}}{{\text{O}}^{+}} \right]$

Complete answer:
We are given that a solution is made by mixing two strong acids – HNO3 and HCl. Their volume and concentration values are also given. To calculate the pH of the resulting solution, we need to follow the given steps:
Step 1. Calculation for the total moles of H+ ions in the solution.
Since HNO3 and Hcl both are strong acids, they will completely dissociate to produce H+ ions. Also, molarity is the number of moles of solute present in 1 liter or 1000 milliliter of solution. Thus, we can calculate number of moles as:

& \text{Molarity }\left( \text{M} \right)=\dfrac{\text{No}\text{. of moles}\left( \text{n} \right)}{\text{Volume in liter}\left( \text{V} \right)} \\\ & \Rightarrow {{\text{M}}_{\text{HCl}}}=\dfrac{{{\text{n}}_{\text{HCl}}}}{{{\text{V}}_{\text{HCl}}}} \\\ & \Rightarrow 0.20\text{ M}=\dfrac{{{\text{n}}_{\text{HCl}}}}{\text{0}\text{.050 mL}} \\\ & \Rightarrow {{\text{n}}_{\text{HCl}}}=0.010\text{ moles} \\\ & \text{Similarly, }{{\text{M}}_{\text{HN}{{\text{O}}_{3}}}}=\dfrac{{{\text{n}}_{\text{HN}{{\text{O}}_{3}}}}}{{{\text{V}}_{\text{HN}{{\text{O}}_{3}}}}} \\\ & \Rightarrow 0.10\text{ M}=\dfrac{{{\text{n}}_{\text{HN}{{\text{O}}_{3}}}}}{\text{0}\text{.1 L}} \\\ & \Rightarrow {{\text{n}}_{\text{HN}{{\text{O}}_{3}}}}=0.010\text{ moles} \\\ & \therefore \text{Total no}\text{. of moles of }{{\text{H}}^{+}}\text{ ions}={{\text{n}}_{\text{HCl}}}+{{\text{n}}_{\text{HN}{{\text{O}}_{3}}}}=0.020\text{ moles} \\\ \end{aligned}$$ Step 2. Calculation of the total volume of the solution in liter. $${{\text{V}}_{\text{Total}}}=100.0\text{mL}+50.0\text{mL}+100.0\text{mL}=250.0\text{mL}$$ Step 3. Calculation of the total concentration of H+ ions in the solution. $$\begin{aligned} & \left[ {{H}_{3}}{{O}^{+}} \right]=\dfrac{\text{Total no}\text{. of moles}}{\text{Total volume in liter}} \\\ & \Rightarrow \left[ {{H}_{3}}{{O}^{+}} \right]=\dfrac{\text{0}\text{.020}}{\text{0}\text{.250}}=0.080\text{ M} \\\ \end{aligned}$$ Step 4. Calculation for the pH of the solution. $$\begin{aligned} & \because \text{pH}=-\log \left[ {{\text{H}}^{+}} \right] \\\ & \therefore \text{pH}=-\log \left[ 0.080 \right]=1.10 \\\ \end{aligned}$$ Hence, the pH of the resulting solution is 1.10. **Note:** A mixture of nitric acid and hydrochloric acid is called aqua regia. Ideally, aqua regia contains nitric acid and hydrochloric acid in a $1:3$ ratio of moles.