Let's translate and solve the problem step-by-step: ### **Translation** "25.00 mL of 0.0500 M HA acid is titrated with 0.2000 M NaOH. At the half-neutralization point (the point where half of the acid is neutralized), the pH is measured as 2.75. Calculate the \(K_a\) value of the acid." --- ### **Step 1: Understanding the Half-Neutralization Point** At the half-neutralization point in a weak acid-strong base titration: - Half of the acid (HA) has been converted to its conjugate base (A⁻). - \([HA] = [A^-]\) - At this point, **pH = pKa** (from the Henderson-Hasselbalch equation). --- ### **Step 2: Write the Relationship** Since **pH = pKa** at the half-neutralization point: \[ \text{pH} = \text{p}K_a \] Given: \[ \text{pH} = 2.75 \implies \text{p}K_a = 2.75 \] --- ### **Step 3: Find \(K_a\) from \(pK_a\)** \[ \text{p}K_a = -\log K_a \] \[ K_a = 10^{-\text{p}K_a} \] \[ K_a = 10^{-2.75} \] --- ### **Step 4: Calculate the Value** \[ K_a = 10^{-2.75} \] \[ K_a \approx 1.78 \times 10^{-3} \] --- ### **Final Answer** **The \(K_a\) value of the acid is:** \[ \boxed{1.78 \times 10^{-3}} \] --- **Explanation:** At the half-neutralization point, pH equals pKa. Using the measured pH (2.75), we directly find pKa and then calculate Ka using the formula \( K_a = 10^{-\text{p}K_a} \).

Let's carefully go through the problem step-by-step with detailed calculations: --- ### **Given Data:** - Volume of acid solution: \( V_{acid} = 25.00\, \text{mL} = 0.025\, \text{L} \) - Concentration of acid (HA): \( M_{acid} = 0.0500\, \text{M} \) - Concentration of NaOH: \( M_{NaOH} = 0.2000\, \text{M} \) - pH at half-neutralization: \( pH = 2.75 \) --- ### **Step 1: Determine the initial moles of acid** Calculate the moles of HA initially present: \[ n_{HA} = M_{acid} \times V_{acid} \] \[ n_{HA} = 0.0500\, \text{mol/L} \times 0.025\, \text{L} = 0.00125\, \text{mol} \] --- ### **Step 2: Understand the half-neutralization point** - At the half-neutralization point: - Half of the acid has reacted. - Moles of acid reacted: \( \frac{n_{HA}}{2} = \frac{0.00125}{2} = 0.000625\, \text{mol} \) - Moles of NaOH added to reach this point: \[ n_{NaOH} = 0.000625\, \text{mol} \] - Volume of NaOH used at this point: \[ V_{NaOH} = \frac{n_{NaOH}}{M_{NaOH}} = \frac{0.000625\, \text{mol}}{0.2000\, \text{mol/L}} = 0.003125\, \text{L} = 3.125\, \text{mL} \] --- ### **Step 3: Calculate the pKa** At the half-neutralization point, the pH equals the pKa: \[ pH = pK_a \] Given: \[ pH = 2.75 \] Therefore: \[ pK_a = 2.75 \] --- ### **Step 4: Calculate \(K_a\)** Using: \[ K_a = 10^{-\text{p}K_a} \] \[ K_a = 10^{-2.75} \] Calculate: \[ K_a = 10^{-2} \times 10^{-0.75} \approx 0.01 \times 10^{-0.75} \] Since: \[ 10^{-0.75} = \frac{1}{10^{0.75}} \approx \frac{1}{5.62} \approx 0.178 \] Thus: \[ K_a \approx 0.01 \times 0.178 = 0.00178 \] or in scientific notation: \[ K_a \approx 1.78 \times 10^{-3} \] --- ### **Final Answer:** \[ \boxed{ \text{The } K_a \text{ of the acid is approximately } \mathbf{1.78 \times 10^{-3}} } \] --- ### **Summary:** - The pH at half-neutralization directly gives pKa. - Using the pKa, we find Ka with \(K_a = 10^{-\text{p}K_a}\). - The calculated \(K_a\) is \(1.78 \times 10^{-3}\). Feel free to ask if you'd like further clarifications!

At the half-neutralization point in a weak acid-strong base titration, pH equals pKa, allowing direct calculation of the acid's dissociation constant \(K_a\).