Buffer Capacity After a Strong-Acid Addition
A 1.00 L aqueous solution at 25 °C contains 0.100 mol of acetic acid (pKa = 4.74) and 0.100 mol of sodium acetate. A technician adds 10.0 mL of 2.00 M HCl with stirring and records the pH before and after. A colleague predicts that, because HCl is a strong acid, the pH will fall by about five units, as it would if the same amount of acid were added to 1.00 L of pure water; the measured fall is less than 0.2 unit. The laboratory protocol states that a single addition may change the pH by no more than 0.20 unit, after which the buffer is treated as exhausted and must be replaced. The colleague's prediction must be checked, the observed pH must be accounted for quantitatively, and the technician wants to know how much more of the same acid could have been added before the protocol limit was reached.
1. Calculate the pH before and after the addition of the HCl, and explain why the observed change is so small.
Read the worked answer
Before addition the solution is an equimolar buffer, so
$$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = 4.74 + \log(1) = 4.74.$$
Moles of HCl added: $0.0100\ \text{L}\times2.00\ \text{mol L}^{-1} = 0.0200$ mol.
The added $\text{H}^+$ is consumed almost completely by the acetate: $$\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$$
- $n(\text{HA})$: $0.100 + 0.0200 = 0.120$ mol
- $n(\text{A}^-)$: $0.100 - 0.0200 = 0.080$ mol
The volume becomes 1.010 L, but the Henderson–Hasselbalch equation uses a ratio, so the volume cancels:
$$\text{pH} = 4.74 + \log\frac{0.080}{0.120} = 4.74 + \log(0.667) = 4.74 - 0.176 = 4.56.$$
Change $= 0.18$ unit, consistent with the measurement and within the 0.20 protocol limit.
The change is small because the strong acid is neutralised by a reservoir of conjugate base: only the ratio of the buffer components changes, and it changes from 1.00 to 0.67, a factor of only 1.5, which is $\log(1.5)\approx0.18$ pH units.
2. Check the colleague's claim by computing the pH change if the same 0.0200 mol of HCl were added to 1.00 L of pure water.
Read the worked answer
In pure water the strong acid is not consumed. After adding 10.0 mL of 2.00 M HCl the total volume is 1.010 L, so
$$[\text{H}^+] = \frac{0.0200}{1.010} = 1.98\times10^{-2}\ \text{mol L}^{-1}$$
$$\text{pH} = -\log(1.98\times10^{-2}) = 1.70.$$
Pure water at 25 °C has pH 7.00, so the fall is
$$7.00 - 1.70 = 5.30\ \text{units}.$$
So the colleague's order of magnitude - 'about five units' - is right for water and wrong for the buffer. The reason is not that HCl is weak but that the buffer supplies $0.100$ mol of acetate, fifty times more than the $0.0200$ mol of $\text{H}^+$ added, so the acid is essentially all converted to acetic acid instead of remaining as free $\text{H}^+$. This is exactly the point of the comparison: the same addition produces a factor of about 30 difference in $[\text{H}^+]$.
3. Estimate how much more 2.00 M HCl could be added before the protocol limit (0.20 unit fall from 4.74) is reached.
Read the worked answer
Set the target pH at $4.74 - 0.20 = 4.54$. Then
$$\log\frac{[\text{A}^-]}{[\text{HA}]} = -0.20 \Rightarrow \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.20} = 0.631.$$
Let $x$ be the total moles of HCl added (including the 0.0200 mol already added):
$$\frac{0.100 - x}{0.100 + x} = 0.631$$
$$0.100 - x = 0.0631 + 0.631x$$
$$0.0369 = 1.631x \Rightarrow x = 0.0226\ \text{mol}.$$
Total volume of 2.00 M HCl corresponding to this: $0.0226/2.00 = 0.0113$ L = 11.3 mL. Since 10.0 mL has already been added, only about 1.3 mL more can be tolerated.
Judgement: the buffer is close to its practical limit despite the pH still looking healthy. As the ratio approaches 0.63, further additions cost progressively more pH change per drop, so the technician should plan to replace the buffer rather than top it up repeatedly. (Worth noting that a buffer works best near a 1:1 ratio; the 0.20-unit protocol limit corresponds to a ratio of about 0.63, i.e. only a modest depletion.)