Questions: MISSED THIS? Watch KCV: The Titration of a Weak Acid and a Strong Base, IWE: Weak Acid-Strong Base Titration pH Curve; Read Section 18.4. You can click on the Review link to access the section in your eText. Consider the titration of a 24.0-mL sample of 0.110 M HC2H3O3(Ka=1.8 x 10^-5) with 0.120 M NaOH. Part A Determine the initial pH. Express your answer to two decimal places. pH=

Solution Steps

The initial concentration of \(\mathrm{H}^+\) ions in the solution can be determined using the acid dissociation constant \(K_a\) for \(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\). The dissociation of \(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\) in water is given by:

\[ \mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2 \leftrightarrow \mathrm{H}^+ + \mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^- \]

The expression for \(K_a\) is:

\[ K_a = \frac{[\mathrm{H}^+][\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]}{[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} \]

Since the initial concentration of \(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\) is 0.110 M and it dissociates slightly, we can assume \( [\mathrm{H}^+] = [\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-] = x \) and \( [\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2] \approx 0.110 - x \approx 0.110 \).

\[ K_a = 1.8 \times 10^{-5} = \frac{x^2}{0.110} \]

Solving for \(x\):

\[ x^2 = 1.8 \times 10^{-5} \times 0.110 \]

\[ x^2 = 1.98 \times 10^{-6} \]

\[ x = \sqrt{1.98 \times 10^{-6}} \]

\[ x = 1.407 \times 10^{-3} \, \text{M} \]

The pH is calculated using the concentration of \(\mathrm{H}^+\) ions:

\[ \mathrm{pH} = -\log [\mathrm{H}^+] \]

\[ \mathrm{pH} = -\log (1.407 \times 10^{-3}) \]

\[ \mathrm{pH} = 2.85 \]

Final Answer