Answer:
0.913
Explanation:
Write the Ka expression for the acetic acid We can get the H+ concentration of the hydrochloric acid by taking the negative log of the molarity since it is a strong acid. From there, we add the concentrations of H+ and take the negative log of it to get the pH of the overall solution.
Solving:
[tex]\subsection*{Strong vs Weak Acid:}\begin{itemize} \item \textbf{\ch{HCl}} is a strong acid and dissociates completely: \[ \ch{HCl} \rightarrow \ch{H+} + \ch{Cl-} \] This means that the \([\ch{H+}]\) from \ch{HCl} is \(0.120 \, \text{M}\). \item \textbf{\ch{HC2H3O2}} (acetic acid) is a weak acid: \[ \ch{HC2H3O2} \rightleftharpoons \ch{H+} + \ch{C2H3O2-} \] The dissociation constant \(K_a\) for acetic acid is given as \(1.8 \times 10^{-5}\).[/tex]
[tex]\subsection*{Using K_a:}Let \(x\) be the amount of \([\ch{H+}]\) dissociated. \[K_a = \frac{[\ch{H+}][\ch{C2H3O2-}]}{[\ch{HC2H3O2}]} = \frac{x \cdot x}{0.250 - x}\]Since \(K_a\) is small, we assume that \(0.250 - x \approx 0.250\), so:\[K_a \approx \frac{x^2}{0.250}\]Now solve for \(x\):\[x^2 = K_a \cdot 0.250 = (1.8 \times 10^{-5}) \cdot 0.250 = 4.5 \times 10^{-6}\]~~~~~~~~~~~~~~~~~~~~~~~~~~x = \sqrt{4.5 \times 10^{-6}} \approx \boxed{2.12 \times 10^{-3} \, \text{M}}\\\][/tex]
[tex]\subsection*{Total \([\ch{H+}]\) Concentration:}\\\\\\\\[[\ch{H+}]_{\text{total}} = [\ch{H+}]_{\ch{HCl}} + [\ch{H+}]_{\ch{HC2H3O2}} = 0.120 \, \text{M} + 2.12 \times 10^{-3} \, \text{M} \approx \boxed{0.1221 \, \text{M}}\]\\[/tex]
[tex]\subsection*{}The pH:\[\text{pH} = -\log [\ch{H+}]\]\[\text{pH} = -\log (0.1221) \approx \boxed{0.913}\][/tex]
Therefore, the pH of the solution is 0.913.