IDNLearn.com: Your trusted source for finding accurate answers. Ask your questions and get detailed, reliable answers from our community of experienced experts.

Determine the pH of the following solution. Initial concentrations are given: [HC₂H₃O₂] = 0.250 M, [HCl] = 0.120 M. [Ka = 1.8 × 10⁻⁵].

Sagot :

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.

We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.