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