Join thousands of students who trust us to help them ace their exams!Watch the first video
Multiple Choice
A benzoate buffer is prepared containing 0.33 M benzoic acid and 0.28 M benzoate. The Ka of benzoic acid is 6.3 x 10^-5. What is the pH of the buffer after the addition of 20 mL of 0.50 M NaOH to 100 mL of this buffer?
A
pH = 4.33
B
pH = 4.19
C
pH = 4.50
D
pH = 4.05
Verified step by step guidance
1
Identify the initial concentrations of benzoic acid (HA) and benzoate (A-) in the buffer. The concentrations are 0.33 M for benzoic acid and 0.28 M for benzoate.
Calculate the moles of benzoic acid and benzoate in the initial 100 mL of buffer. Use the formula: moles = concentration (M) x volume (L).
Determine the moles of NaOH added. Use the formula: moles = concentration (M) x volume (L) for the 20 mL of 0.50 M NaOH.
Use the stoichiometry of the reaction between NaOH and benzoic acid to find the new concentrations. NaOH will react with benzoic acid to form water and benzoate, changing their concentrations.
Apply the Henderson-Hasselbalch equation to find the new pH of the buffer. The equation is: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \( \text{pK}_a = -\log(\text{K}_a) \).