Multiple Choice
Calculate the pH of a 0.300 M NaHCOO solution, given that the Ka for HCOOH is 1.8 x 10^-4.
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the percent ionization of a 0.16 M benzoic acid solution in a solution containing 0.28 M sodium benzoate, given that the Ka of benzoic acid is 6.3 x 10^-5?
A
0.25%
B
5.0%
C
3.2%
D
1.5%
Verified step by step guidance1
Identify the components of the solution: benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa). Benzoic acid is a weak acid, and sodium benzoate is its conjugate base.
Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \([\text{A}^-]\) is the concentration of the conjugate base (sodium benzoate) and \([\text{HA}]\) is the concentration of the weak acid (benzoic acid).
Calculate the pKa from the given Ka: \( \text{pKa} = -\log(6.3 \times 10^{-5}) \).
Substitute the concentrations of benzoic acid and sodium benzoate into the Henderson-Hasselbalch equation to calculate the pH.
Determine the percent ionization using the formula: \( \text{Percent Ionization} = \left( \frac{[\text{H}^+]}{[\text{HA}]} \right) \times 100 \% \), where \([\text{H}^+]\) is the concentration of hydrogen ions, which can be found from the pH.
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