9. Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
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An unknown diprotic acid has an initial concentration of 0.025 M. What is the pH of the solution if pka1 is 3.25 and pKa2 is 6.82?
A
4.21
B
2.46
C
1.91
D
3.25
E
2.79
Verified step by step guidance1
Identify that the problem involves a diprotic acid, which can donate two protons (H⁺ ions) in solution. This means it has two dissociation steps, each with its own equilibrium constant (Ka1 and Ka2).
Use the given pKa1 value to find Ka1. Recall that pKa is the negative logarithm of the acid dissociation constant: \( \text{pKa1} = -\log(Ka1) \). Therefore, \( Ka1 = 10^{-\text{pKa1}} \).
Recognize that for a diprotic acid, the first dissociation step is the most significant in determining the initial pH, especially when the initial concentration is relatively low. Use the expression for the first dissociation: \( HA \rightleftharpoons H^+ + A^- \).
Set up the equilibrium expression for the first dissociation: \( Ka1 = \frac{[H^+][A^-]}{[HA]} \). Assume that the concentration of \( H^+ \) and \( A^- \) at equilibrium is \( x \), and the concentration of \( HA \) is \( 0.025 - x \).
Solve the equilibrium expression for \( x \), which represents the concentration of \( H^+ \) ions. Then, calculate the pH using the formula \( \text{pH} = -\log[H^+] \).
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