# Question #14369

May 7, 2015

The pH of the resulting solution will be 1.

So, you're mixing three hydrochloric acid solutions, each with a different molarity and volume. In order to get the pH of the resulting solution, you need to first determine what the molarity of the resulting solution will be.

Since molarity is defined as moles of solute per liters of solution, you can determine how many moles of $\text{HCl}$ you add with each solution

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{sol 1" = 50 * 10^(-3)"L" * "0.2 M" = "0.01 moles HCl}}$

${n}_{\text{sol 2" = 50 * 10^(-3)"L" * "0.1 M" = "0.005 moles HCL}}$

${n}_{\text{sol3" = 100 * 10^(-3)"L" * "0.05 M" = "0.05 moles HCl}}$

The total number of moles of hydrochloric acid will be

${n}_{\text{total" = n_"sol 1" + n_"sol 2" + n_"sol 3}}$

${n}_{\text{total" = 0.01 + 0.005 + 0.005 = "0.02 moles HCl}}$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{1} + {V}_{2} + {V}_{3}$

${V}_{\text{total" = 50 + 50 + 100 = "200 mL}}$

This means that the molarity of the final solution will be

$C = \frac{n}{V} = \text{0.02 moles"/(200 * 10^(-3)"L") = "0.1 M}$

You know that

$H C {l}_{\left(a q\right)} \to {H}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

Because hydrochloric acid dissociates completely, the concentration of the ${H}^{+}$ ions will also be equal to 0.1 M.

As a result, the pH of the solution will be

$p {H}_{\text{sol}} = - \log \left(\left[{H}^{+}\right]\right)$

$p {H}_{\text{sol}} = - \log \left(0.1\right) = \textcolor{g r e e n}{1}$