Buffer capacity: how much HCl before pH 4.75 collapses?
A school laboratory must supply 250.0 mL of a buffer of pH 4.75 for a project on enzyme activity. The shelf holds only two reagent bottles: 0.100 M acetic acid and 0.100 M sodium acetate solution. For acetic acid, Ka = 1.8 × 10⁻⁵, so pKa = 4.74. A student mixes 125.0 mL of each, checks the pH meter and finds 4.74, and decides that the difference "does not matter". To demonstrate the value of buffering, the teacher takes two beakers: beaker X with 100.0 mL of the buffer prepared above, and beaker Y with 100.0 mL of distilled water at pH 7.00. A 1.00 M HCl solution is added to each beaker in 2.00 mL portions with stirring, and the pH is read after each addition. The water's pH falls to 1.7 after the very first portion. A visitor remarks that since a buffer "resists pH change", the teacher can keep adding HCl indefinitely without the pH changing much.
1. Check the student's volumes: is 125.0 mL + 125.0 mL the correct recipe for pH 4.75? Judge whether accepting the meter reading of 4.74 was reasonable.
Read the worked answer
Use pH = pKa + log([A⁻]/[HA]).
4.75 − 4.74 = 0.01, so [A⁻]/[HA] = 10^0.01 = 1.023.
With equal stock concentrations, volumes follow the same ratio. Let V(HA) + V(A⁻) = 250.0 mL and V(A⁻)/V(HA) = 1.023:
V(HA) = 250.0/2.023 = 123.6 mL, V(A⁻) = 126.4 mL.
So 125 + 125 is not the exact recipe. But the error is tiny: at a 1:1 ratio the pH equals pKa = 4.74, only 0.01 unit below target, which is below the reproducibility of a school pH meter (±0.05). Accepting 4.74 is therefore reasonable - but the reasoning should be "the ratio is 1.02, so the deviation is within meter error", not "the difference does not matter". The second statement would justify any sloppy ratio.
2. Compute the pH of beaker X after the first 2.00 mL portion of 1.00 M HCl, compare it with beaker Y, and explain the difference quantitatively.
Read the worked answer
100.0 mL of the buffer contains 0.0500 mol/L × 0.100 L = 5.00 mmol HA and 5.00 mmol A⁻.
First portion adds 2.00 mL × 1.00 M = 2.00 mmol H⁺, which converts acetate to acetic acid:
- A⁻ = 5.00 − 2.00 = 3.00 mmol
- HA = 5.00 + 2.00 = 7.00 mmol
pH = 4.74 + log(3.00/7.00) = 4.74 − 0.37 = 4.37 (a fall of 0.37 unit).
Beaker Y: 2.00 mmol in 102 mL gives [H⁺] = 0.0196 M, pH = 1.71, a fall of 5.3 units.
The buffer's conjugate ratio moved only from 1.00 to 0.43 - a factor of 2.3 - and the logarithm compresses even that into 0.37 unit. Water has no reservoir of base at all, so all the added H⁺ stays free.
3. Track the later portions: at which point does the buffer stop working, and what is the pH after the third portion? Was the visitor right?
Read the worked answer
Each portion delivers 2.00 mmol H⁺, but the beaker holds only 5.00 mmol of acetate, so it can neutralise 2.5 portions.
After the 2nd portion (4.00 mmol added): A⁻ = 1.00 mmol, HA = 9.00 mmol pH = 4.74 + log(1/9) = 4.74 − 0.954 = 3.79 - the buffer is being stretched but is still functioning.
After the 3rd portion (6.00 mmol added): all 5.00 mmol of A⁻ is consumed and 1.00 mmol H⁺ remains in 106 mL. [H⁺] = 1.00 × 10⁻³/0.106 = 9.4 × 10⁻³ M → pH ≈ 2.03. The 0.094 M acetic acid formed contributes only about 1.3 × 10⁻³ M H⁺, so the strong acid dominates; the drop is essentially that of an unbuffered solution.
So the buffer fails between the 2nd and 3rd portions. Capacity is fixed by the amount of A⁻ present, not by the label "buffer". The visitor is wrong: resistance to pH change is finite, and can be calculated in advance.