An analytical chemist is titrating 56.7 mL of a 0.4900M solution of formic acid (H,CO,) with a 0.7500M solution of KOH. The p K, of formic acid is 3.74. Calculate the pH of the acid solution after the chemist has added 44.2 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. pH =

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 114CWP: Consider the following acids and bases: HCO2H Ka = 1.8 104 HOBr Ka = 2.0 109 (C2H5)2NH Kb = 1.3 ...
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### Titration Problem on Calculating pH

**Problem Statement:**

An analytical chemist is titrating 56.7 mL of a 0.4900 M solution of formic acid (\(H_2CO_2\)) with a 0.7500 M solution of KOH. The \(pK_a\) of formic acid is 3.74. Calculate the pH of the acid solution after the chemist has added 44.2 mL of the KOH solution to it.

**Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.

Round your answer to 2 decimal places.

**Calculation:**

1. **Determine moles of formic acid and KOH:**

   - Moles of formic acid = \(0.4900 \, M \times 0.0567 \, L = 0.027783 \, moles\)
   - Moles of KOH = \(0.7500 \, M \times 0.0442 \, L = 0.03315 \, moles\)

2. **Neutralization:**

   Since \(KOH\) is a strong base, it will neutralize an equivalent amount of formic acid. The moles of formic acid will decrease by the number of moles of \(KOH\) added:

   - Remaining moles of formic acid = \(0.027783 \, moles - 0.03315 \, moles = -0.005367 \, moles\)

   Since the moles of \(KOH\) exceed the moles of formic acid, formic acid is completely neutralized and excess \(KOH\) remains in the solution.

3. **Calculating the concentration of \(OH^-\) ions:**

   - Excess moles of \(KOH\) = \(0.03315 \, moles - 0.027783 \, moles = 0.005367 \, moles\)
   - Total volume of the solution = \(56.7\, mL + 44.2\, mL = 100.9\, mL = 0.1009\, L\)
   - Concentration of \(KOH\) = \(\frac{0.005367 \, moles
Transcribed Image Text:### Titration Problem on Calculating pH **Problem Statement:** An analytical chemist is titrating 56.7 mL of a 0.4900 M solution of formic acid (\(H_2CO_2\)) with a 0.7500 M solution of KOH. The \(pK_a\) of formic acid is 3.74. Calculate the pH of the acid solution after the chemist has added 44.2 mL of the KOH solution to it. **Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. **Calculation:** 1. **Determine moles of formic acid and KOH:** - Moles of formic acid = \(0.4900 \, M \times 0.0567 \, L = 0.027783 \, moles\) - Moles of KOH = \(0.7500 \, M \times 0.0442 \, L = 0.03315 \, moles\) 2. **Neutralization:** Since \(KOH\) is a strong base, it will neutralize an equivalent amount of formic acid. The moles of formic acid will decrease by the number of moles of \(KOH\) added: - Remaining moles of formic acid = \(0.027783 \, moles - 0.03315 \, moles = -0.005367 \, moles\) Since the moles of \(KOH\) exceed the moles of formic acid, formic acid is completely neutralized and excess \(KOH\) remains in the solution. 3. **Calculating the concentration of \(OH^-\) ions:** - Excess moles of \(KOH\) = \(0.03315 \, moles - 0.027783 \, moles = 0.005367 \, moles\) - Total volume of the solution = \(56.7\, mL + 44.2\, mL = 100.9\, mL = 0.1009\, L\) - Concentration of \(KOH\) = \(\frac{0.005367 \, moles
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