The Henderson-Hasselbalch equation for a buffer solution is:
pH = pKa + log([conjugate base]/[weak acid])
where pKa is the acid dissociation constant, [conjugate base] is the concentration of the conjugate base, and [weak acid] is the concentration of the weak acid.
First, we need to find the pKa for benzoic acid using the given equilibrium constant (K):
K = [H+][C6H5COO-]/[HC6H5O2]
K = (x)(0.060)/0.020
K = 0.18x
Assuming x << 0.020, we can simplify the equation:
K = (x)(0.060)/0.020 ≈ (x)(0.060)/0.020 - x
0.18x ≈ 0.060x/0.020 - x
0.18x ≈ 3x - x
0.18x ≈ 2x
x ≈ 0.011
Therefore, [H+] = 0.011 M
pH = -log(0.011) = 1.96
Next, we need to calculate the concentrations of benzoic acid and sodium benzoate in the buffer solution:
moles of benzoic acid = 1.00 L x 0.020 mol/L = 0.020 mol
moles of sodium benzoate = 3.00 L x 0.060 mol/L = 0.180 mol
concentration of benzoic acid = 0.020 mol/4.00 L = 0.005 M
concentration of sodium benzoate = 0.180 mol/4.00 L = 0.045 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log([conjugate base]/[weak acid])
pH = 4.80 + log(0.045/0.005)
pH = 4.80 + 0.69
pH = 5.49
Therefore, the pH of the buffer solution is 5.49.
What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC,H,O,, with 3.00 L of 0.060 M sodium benzoate,
NaC HO₂? K for benzoic acid is 6.3 x 10-5.
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