Answer:
pH = 8.11
Explanation:
Step 1: Data given
Volume of a 0.150 M benzoic acid sample = 25.0 mL
Molarity of NaOH solution = 0.150 M
Ka of the benzoic acid = 4.50 * 10^-4
Step 2: The balanced equation
C7H6O2 + NaOH → C7H5O2Na + H2O
Step 3: calculate moles
At the equivalence point we have the same amount of of acid and base
Both benzoic acid and NaOH will completely be consumed.
Moles benzoic acid = molarity *volume
Moles benzoic acid = 0.150 M * 0.025 L
Moles benzoic acid = 0.00375 moles
We we'll need 0.00375 moles NaOH to reach the equivalence point
Volume NaOH = 0.00375 moles / 0.150 M
Volume = 0.025 L = 25 mL
Total volume = 25 + 25 = 50 mL
Step 4: Calculate moles of C7H5O2Na
For 1 mol benzoic acid we need 1 mol NaOH to produce 1 mol C7H5O2Na
For 0.00375 moles benzoic acid we'll have 0.00375 moles C7H5O2Na
Step 5: The equation
C7H5O2- + H2O → C7H6O2 + OH-
Step 6: The initial concentration
[C7H5O2-] = 0.00375 moles / 0.050 L
[C7H5O2-] = 0.075 M
[C7H6O2] = 0M
[OH-] = 0M
Step 7: The concentration at the equilibrium
[C7H5O2-] = 0.075 - X M
[C7H6O2] = XM
[OH-] = XM
Step 8: Calculate Kb
Ka * Kb = Kw
Kb = Kw / Ka
Kb = 10^-14 / (4.50 *10^-4)
Kb = 2.22 *10^-11
Kb = [C7H6O2][OH-]/[C7H5O2-]
2.22 * 10^-11 = X²/ 0.075 - X
2.22 * 10^-11 = X²/ 0.075
X² = 2.22 *10^-11 * 0.075
X² = 1.665 *10^-12
X = 1.29 * 10^-6 = [OH-]
Step 9: Calculate pOH
pOH = -log[OH-]
pOH = -log(1.29 * 10^-6 )
pOH = 5.89
Step 10: Calculate pH
pH = 14 - 5.89
pH = 8.11
