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Ch.17 - Applications of Aqueous Equilibria
Chapter 17, Problem 96

Consider the titration of 25.0 mL of 0.0200 M H2CO3 with 0.0250 M KOH. Calculate the pH after the addition of each of the following volumes of base: (a) 10.0 mL, (b) 20.0 mL, (c) 30.0 mL, (d) 40.0 mL.

Verified step by step guidance
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Step 1: Determine the initial moles of H2CO3. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Calculate the moles of H2CO3 in the initial 25.0 mL solution.
Step 2: Calculate the moles of KOH added for each volume. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Perform this calculation for 10.0 mL, 20.0 mL, 30.0 mL, and 40.0 mL of KOH.
Step 3: For each volume of KOH added, determine the reaction progress. H2CO3 reacts with KOH in a 1:1 molar ratio to form HCO3^- and water. Subtract the moles of KOH from the initial moles of H2CO3 to find the remaining moles of H2CO3 and the moles of HCO3^- formed.
Step 4: Calculate the pH for each scenario. For the buffer region (before equivalence point), use the Henderson-Hasselbalch equation: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \right) \). For the equivalence point and beyond, consider the hydrolysis of HCO3^- or excess OH^- to find the pH.
Step 5: Analyze the results. Compare the calculated pH values for each volume of KOH added to understand the titration curve and the buffering capacity of the solution.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration

Titration is a quantitative analytical technique used to determine the concentration of a solute in a solution. It involves the gradual addition of a titrant (in this case, KOH) to a solution of the analyte (H2CO3) until the reaction reaches its equivalence point, where stoichiometrically equivalent amounts of acid and base have reacted. The pH changes during this process can be monitored to determine the endpoint of the titration.
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pH and Acid-Base Chemistry

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. In acid-base chemistry, the pH changes as acids and bases react, and understanding how to calculate pH at various points in a titration is crucial. For weak acids like H2CO3, the pH will vary depending on the amount of base added and the degree of dissociation of the acid.
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Buffer Solutions

Buffer solutions are mixtures that resist changes in pH upon the addition of small amounts of acid or base. In the context of this titration, the presence of H2CO3 and its conjugate base (HCO3-) can create a buffer system that stabilizes the pH in certain regions of the titration curve. Understanding how buffers work is essential for predicting pH changes at different volumes of KOH added.
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Related Practice
Textbook Question

Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H2A+: Ka1 = 4.6 × 10–3, Ka2 = 2.0 × 10–10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (b) 25.0 mL (c) 50.0 mL

Textbook Question

Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H2A+: Ka1 = 4.6 × 10–3, Ka2 = 2.0 × 10–10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (d) 75.0 mL

Textbook Question

Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H2A+: Ka1 = 4.6 × 10–3, Ka2 = 2.0 × 10–10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (e) 100.0 mL

Textbook Question

Consider the titration of 50.0 mL of 1.00 M H3PO4 with 1.00 M KOH. Calculate the pH after the addition of each of the following volumes of base. (a) 25.0 mL (c) 75.0 mL

Textbook Question

Consider the titration of 50.0 mL of 1.00 M H3PO4 with 1.00 M KOH. Calculate the pH after the addition of each of the following volumes of base. (b) 50.0 mL (d) 100.0 mL

Textbook Question
The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equiva-lence point and 9.27 at the second equivalence point. If the acid solution contained 0.2015 g of the acid, what is the molar mass, pKa1, and pKa2 of the acid?