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.
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.
To find the pH of the buffer, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the buffer
pKa = the acid dissociation constant of the weak acid in the buffer system
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, benzoic acid (HC7H5O2) is a weak acid, and sodium benzoate (NaC7H5O2) is its conjugate base.
First, let's calculate the concentration of both the weak acid and its conjugate base after mixing them.
For the weak acid (benzoic acid):
Initial concentration = 0.020 M
Dilution factor = 1.00 L / (1.00 L + 3.00 L) = 0.250
Concentration after dilution = 0.020 M * 0.250 = 0.005 M
For the conjugate base (sodium benzoate):
Initial concentration = 0.060 M
Dilution factor = 3.00 L / (1.00 L + 3.00 L) = 0.750
Concentration after dilution = 0.060 M * 0.750 = 0.045 M
Now, let's calculate the pKa value for benzoic acid:
pKa = -log10(Ka)
Given that Ka is 6.3 x 10^(-5), we can calculate pKa:
pKa = -log10(6.3 x 10^(-5))
pKa ≈ 4.2
Now, we can substitute the values into the Henderson-Hasselbalch equation to find the pH:
pH = 4.2 + log(0.045/0.005)
pH ≈ 4.2 + log(9)
pH ≈ 4.2 + 0.954
pH ≈ 5.154
Therefore, the pH of the buffer is approximately 5.154.