Textbook Question
A 100.0 mL sample of a solution that is 0.100 M in HCl and 0.100 M in HCN is titrated with 0.100 M NaOH. Calculate the pH after the addition of the following volumes of NaOH: (b) 75.0 mL
Ch.17 - Applications of Aqueous Equilibria
Chapter 17, Problem 146
When a typical diprotic acid H2A (Ka1 = 10^-4, Ka2 = 10^-10) is titrated with NaOH, the principal A-containing species at the first equivalence point is HA-. (b) Assuming that this is the principal reaction, show that the pH at the first equivalence point equals the average of pKa1 and pKa2. (c) How many A2- ions are present in 50.0 mL of 1.0 M NaHA?
Verified step by step guidance1
Step 1: Understand the concept of equivalence point in a titration. At the first equivalence point of a diprotic acid titration, all the initial H2A has been converted to HA-. This is because the amount of NaOH added is stoichiometrically equivalent to the amount of H2A initially present.
Step 2: Calculate the pKa values from the given Ka values. Use the formula \( \text{pKa} = -\log(\text{Ka}) \) to find pKa1 and pKa2. For Ka1 = 10^{-4}, pKa1 = 4. For Ka2 = 10^{-10}, pKa2 = 10.
Step 3: Determine the pH at the first equivalence point. At this point, the solution contains HA- ions, which can act as both an acid and a base. The pH is approximately the average of pKa1 and pKa2 because HA- is a weak acid and a weak base, and the solution is a buffer. Calculate the average: \( \text{pH} = \frac{\text{pKa1} + \text{pKa2}}{2} \).
Step 4: Calculate the number of A^{2-} ions in 50.0 mL of 1.0 M NaHA. First, determine the number of moles of NaHA: \( \text{moles of NaHA} = \text{volume (L)} \times \text{molarity} = 0.050 \times 1.0 \).
Step 5: Recognize that in a solution of NaHA, the HA- ions can dissociate to form A^{2-} ions. However, the extent of this dissociation is determined by the second dissociation constant, Ka2. Use the equilibrium expression for the dissociation of HA- to calculate the concentration of A^{2-} ions.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Diprotic Acids and Their Ionization
Diprotic acids can donate two protons (H+) in a stepwise manner, characterized by two dissociation constants, Ka1 and Ka2. The first ionization produces a conjugate base (HA-) and the second ionization produces A2-. Understanding these ionization steps is crucial for analyzing the behavior of diprotic acids during titration.
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04:41
3 forms of Diprotic Acids
Equivalence Point in Titration
The equivalence point in a titration is reached when the amount of titrant added is stoichiometrically equivalent to the amount of analyte present. For diprotic acids, the first equivalence point occurs when half of the acid has been converted to its first conjugate base (HA-), which is essential for calculating pH and understanding the species present at this stage.
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03:06Equivalence Point in Titration
pH and pKa Relationship
The pH at the equivalence point of a diprotic acid titration can be determined using the average of the pKa values of the acid. This relationship arises from the Henderson-Hasselbalch equation, which relates pH, pKa, and the ratio of the concentrations of the acid and its conjugate base, providing a way to predict the pH based on the acid's dissociation constants.
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02:09pH and pOH Calculations
Related Practice
Textbook Question
A 0.0100 mol sample of solid Cd(OH)2 (Ksp = 5.3 x 10^-15) in 100.0 mL of water is titrated with 0.100 M HNO3. (c) How many milliliters of 0.100 M HNO3 must be added to completely neutralize the Cd(OH)2?
Textbook Question
Ethylenediamine (NH2CH2CH2NH2, abbreviated en) is an organic base that can accept two protons: (a) Consider the titration of 30.0 mL of 0.100 M ethylenediamine with 0.100 M HCl. Calculate the pH after the addition of the following volumes of acid, and construct a qualitative plot of pH versus milliliters of HCl added: (i) 0.0 mL (ii) 15.0 mL (iii) 30.0 mL (iv) 45.0 mL (v) 60.0 mL (vi) 75.0 mL
Textbook Question
A 40.0 mL sample of a mixture of HCl and H3PO4 was titrated with 0.100 M NaOH. The first equivalence point was reached after 88.0 mL of base, and the second equivalence point was reached after 126.4 mL of base. (a) What is the concentration of H3O+ at the first equivalence point?