32. A buffer solution is composed of 0.1 M acetic acid and 0.15 M sodium acetate. The change in pH of 1 L buffer solution upon addition of 50 mL of 1.0 M NaOH is _____, rounded off to two decimal places. The Ka of acetic acid is 1.75 × 10−5.
How to Calculate the Change in pH of a Buffer After Adding NaOH
Correct Answer: 0.43
The change in pH of the buffer after adding NaOH is 0.43 pH unit. The given buffer contains acetic acid, CH3COOH, as the weak acid and sodium acetate, CH3COONa, as the source of its conjugate base, CH3COO−. When NaOH is added, the strong base does not simply remain as free OH− in the buffer. Instead, it reacts almost completely with acetic acid and converts part of it into acetate ion.
Therefore, the correct method is to first perform the neutralization stoichiometry and calculate the new amounts of acetic acid and acetate ion. The Henderson–Hasselbalch equation can then be applied before and after NaOH addition. The initial pH is approximately 4.93, while the final pH is approximately 5.36. Hence, the increase in pH is approximately 0.43.
Understanding the Acetic Acid–Sodium Acetate Buffer
What Components Are Present in the Buffer?
The buffer contains a weak acid and its conjugate base. Acetic acid, CH3COOH, is the weak acid, while sodium acetate dissociates completely in water to provide acetate ions, CH3COO−.
The relevant acid dissociation equilibrium is:
CH3COOH ⇌ H+ + CH3COO−
The pH of such a weak acid–conjugate base buffer is calculated using the Henderson–Hasselbalch equation:
pH = pKa + log([CH3COO−]/[CH3COOH])
The ratio of conjugate base to weak acid determines how far the buffer pH lies above or below the pKa of the weak acid.
Step-by-Step Solution of the Buffer pH Problem
Step 1: Calculate the Initial Moles of Acetic Acid
The volume of the original buffer solution is 1 L, and the concentration of acetic acid is 0.1 M. The number of moles is calculated from molarity and volume.
Moles = Molarity × Volume in litres
Therefore:
Moles of CH3COOH = 0.1 × 1 = 0.10 mol
Thus, the original buffer contains 0.10 mol of acetic acid.
Step 2: Calculate the Initial Moles of Acetate Ion
The concentration of sodium acetate is 0.15 M in the same 1 L buffer solution. Since sodium acetate dissociates to provide one mole of acetate ion per mole of sodium acetate:
Moles of CH3COO− = 0.15 × 1 = 0.15 mol
Therefore, before the addition of NaOH, the buffer contains:
CH3COOH = 0.10 mol
CH3COO− = 0.15 mol
Step 3: Calculate the pKa of Acetic Acid
The acid dissociation constant of acetic acid is given as:
Ka = 1.75 × 10−5
The pKa is calculated using:
pKa = −log Ka
Therefore:
pKa = −log(1.75 × 10−5)
pKa ≈ 4.76
Step 4: Calculate the Initial pH of the Buffer
Using the Henderson–Hasselbalch equation:
pHinitial = pKa + log([A−]/[HA])
Since both buffer components are present in the same volume, their concentration ratio is equal to their mole ratio. Therefore:
pHinitial = 4.76 + log(0.15/0.10)
pHinitial = 4.76 + log(1.5)
Since:
log(1.5) ≈ 0.1761
Therefore:
pHinitial ≈ 4.76 + 0.1761 = 4.93
Thus, the initial pH of the buffer is approximately 4.93.
Effect of Adding NaOH to the Buffer
Step 5: Calculate the Moles of NaOH Added
The added NaOH solution has a concentration of 1.0 M and a volume of 50 mL. First, the volume is converted into litres:
50 mL = 0.050 L
The number of moles of NaOH added is:
Moles of NaOH = 1.0 × 0.050
Moles of NaOH = 0.050 mol
Since NaOH is a strong base, it dissociates completely and provides 0.050 mol of OH−.
Step 6: Write the Neutralization Reaction
The added hydroxide ions react with acetic acid according to the equation:
CH3COOH + OH− → CH3COO− + H2O
The reaction occurs in a 1:1 molar ratio. Therefore, 0.050 mol of OH− consumes exactly 0.050 mol of acetic acid and simultaneously forms 0.050 mol of acetate ion.
Step 7: Calculate the Moles of Acetic Acid Remaining
Initially, the buffer contains 0.10 mol of acetic acid. The added NaOH consumes 0.050 mol.
Final moles of CH3COOH = 0.10 − 0.050
Final moles of CH3COOH = 0.050 mol
Therefore, half of the original acetic acid remains after the neutralization reaction.
Step 8: Calculate the Final Moles of Acetate Ion
Initially, the buffer contains 0.15 mol of acetate ion. Neutralization of 0.050 mol of acetic acid produces an additional 0.050 mol of acetate.
Final moles of CH3COO− = 0.15 + 0.050
Final moles of CH3COO− = 0.20 mol
The buffer composition after adding NaOH is therefore:
CH3COOH = 0.050 mol
CH3COO− = 0.20 mol
Calculation of the Final Buffer pH
Step 9: Apply the Henderson–Hasselbalch Equation After Neutralization
The final pH is calculated using the new amounts of weak acid and conjugate base:
pHfinal = pKa + log(nA−/nHA)
Substituting the values:
pHfinal = 4.76 + log(0.20/0.050)
pHfinal = 4.76 + log(4)
Since:
log(4) ≈ 0.6021
Therefore:
pHfinal ≈ 4.76 + 0.6021 = 5.36
Thus, after adding NaOH, the pH of the buffer increases to approximately 5.36.
Step 10: Calculate the Change in pH
The change in pH is calculated by subtracting the initial pH from the final pH:
ΔpH = pHfinal − pHinitial
ΔpH = 5.36 − 4.93
ΔpH ≈ 0.43
Therefore, the pH increases by approximately 0.43 pH unit.
Direct Calculation of the Change in pH
Since the same weak acid and the same Ka are involved before and after adding NaOH, the pKa terms cancel when the change in pH is calculated. Therefore, the answer can be obtained more directly from the change in the conjugate base-to-acid ratio.
Initially:
[A−]/[HA] = 0.15/0.10 = 1.5
After adding NaOH:
[A−]/[HA] = 0.20/0.050 = 4
Therefore:
ΔpH = log(4) − log(1.5)
Using the logarithm rule:
ΔpH = log(4/1.5)
ΔpH = log(2.6667)
ΔpH ≈ 0.426
Rounded off to two decimal places:
ΔpH = 0.43
Why the Increase in Total Volume Does Not Affect the Final Ratio
After adding 50 mL of NaOH solution to 1 L of buffer, the total volume becomes approximately 1.05 L. However, the Henderson–Hasselbalch equation depends on the ratio of conjugate base concentration to weak acid concentration.
Both species are present in the same final volume. Therefore:
[A−]/[HA] = (nA−/V)/(nHA/V)
The common volume cancels:
[A−]/[HA] = nA−/nHA
For this reason, the final mole ratio 0.20/0.050 can be used directly without separately dividing both quantities by 1.05 L.
Why NaOH Must Be Neutralized Before Using the Buffer Equation
The Henderson–Hasselbalch equation should not be applied immediately using the original buffer concentrations after a strong base has been added. NaOH reacts chemically with the weak acid component of the buffer before a new acid–base equilibrium is established.
In this problem, 0.050 mol of NaOH converts 0.050 mol of acetic acid into 0.050 mol of acetate ion. Therefore, the weak acid decreases from 0.10 mol to 0.050 mol, while the conjugate base increases from 0.15 mol to 0.20 mol.
Only after completing this neutralization calculation should the Henderson–Hasselbalch equation be used with the new buffer composition.
How the Buffer Resists the Added Strong Base
A buffer resists a sudden change in pH because one of its components reacts with an added acid or base. When NaOH is introduced into the acetic acid–acetate buffer, the acetic acid component removes the added OH− ions by converting them into water.
The reaction is:
CH3COOH + OH− → CH3COO− + H2O
As a result, the strong base is transformed into the much weaker conjugate base, acetate ion. The pH still increases because the ratio of acetate to acetic acid becomes larger, but the buffer prevents the added OH− from remaining freely in the solution.
Complete Calculation in One Expression
The change in pH can be calculated directly as:
ΔpH = log[(0.15 + 0.05)/(0.10 − 0.05)] − log(0.15/0.10)
ΔpH = log(0.20/0.05) − log(1.5)
ΔpH = log(4) − log(1.5)
ΔpH = log(2.6667)
ΔpH = 0.426 ≈ 0.43
Final Answer
Initially, 1 L of the buffer contains 0.10 mol of acetic acid and 0.15 mol of acetate ion. The addition of 50 mL of 1.0 M NaOH introduces 0.050 mol of OH−, which reacts completely with 0.050 mol of acetic acid. After neutralization, 0.050 mol of acetic acid remains and the amount of acetate ion increases to 0.20 mol.
The initial buffer ratio is 0.15/0.10 = 1.5, while the final buffer ratio is 0.20/0.050 = 4. Using the Henderson–Hasselbalch equation, the pH change is log(4/1.5) = 0.426.
Therefore, the change in pH of the buffer solution, rounded off to two decimal places, is 0.43.
Correct Answer: 0.43
