Multiple Choice
If 17.0 mL of glacial acetic acid (pure HC2H3O2; Ka=1.8×10⁻⁵) is diluted to 1.65 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.
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17. Acid and Base Equilibrium
pH of Weak Acids
Multiple Choice
What is the pH of a solution that is 0.060 M in potassium propionate and 0.085 M in propionic acid, given that the Ka of propionic acid is 1.3 x 10^-5?
A
3.92
B
6.02
C
5.13
D
4.87
Verified step by step guidance1
Identify the components of the solution: potassium propionate (a salt) and propionic acid (a weak acid). This is a buffer solution, which resists changes in pH.
Use the Henderson-Hasselbalch equation to find the pH of the buffer solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \([\text{A}^-]\) is the concentration of the conjugate base (potassium propionate), and \([\text{HA}]\) is the concentration of the weak acid (propionic acid).
Calculate the pKa from the given Ka of propionic acid: \( \text{pKa} = -\log(\text{Ka}) \). Substitute \( \text{Ka} = 1.3 \times 10^{-5} \) into the equation to find \( \text{pKa} \).
Substitute the concentrations of potassium propionate (0.060 M) and propionic acid (0.085 M) into the Henderson-Hasselbalch equation: \( \text{pH} = \text{pKa} + \log \left( \frac{0.060}{0.085} \right) \).
Calculate the logarithmic term \( \log \left( \frac{0.060}{0.085} \right) \) and add it to the pKa to find the pH of the solution.
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