A student prepares a buffer from 0.50 M acetic acid and 0.50 M sodium acetate (pKa = 4.74), then dilutes the entire solution to twice its volume with pure water. What happens to the pH and the buffer capacity?
ApH increases and capacity stays the same, because dilution increases the proportion of the basic acetate component
BpH stays approximately the same but capacity decreases, because the [A⁻]/[HA] ratio is unchanged but fewer total moles of conjugate pair remain
CpH decreases because dilution shifts the acid-base equilibrium toward greater dissociation of HA
DBoth pH and capacity remain exactly the same because a buffer resists all changes, including dilution with water
Henderson-Hasselbalch shows pH depends on the logarithm of [A⁻]/[HA]. Diluting both components by the same factor leaves their ratio unchanged, so the log term is unchanged and pH barely moves. Buffer capacity, however, depends on the total moles of conjugate pair available to absorb acid or base — with half as many moles, the buffer exhausts faster. Dilution preserves pH but reduces capacity. Option 3 is the dangerous misconception: buffers resist pH changes from added acid or base, not changes in the number of available moles.
Question 2 Multiple Choice
When a small amount of hydrochloric acid is added to a buffer containing acetic acid (HA) and acetate (A⁻), what prevents the pH from dropping sharply?
AThe acetic acid reacts with the added H⁺ to regenerate water and acetate, consuming the acid
BThe buffer dilutes the added H⁺ across the large solution volume, reducing its effective concentration
CThe acetate (A⁻) reacts with the added H⁺ to form acetic acid (HA), neutralizing the free proton before it can lower pH
DBoth HA and A⁻ react proportionally with H⁺, maintaining the ratio between them
The acetate ion is the basic component of the buffer. When H⁺ is added, A⁻ captures it: A⁻ + H⁺ → HA. This converts one buffer component into the other without allowing free H⁺ to accumulate. The ratio [A⁻]/[HA] shifts slightly (less A⁻, more HA), which nudges pH slightly per Henderson-Hasselbalch — but the change is far smaller than adding H⁺ to pure water. Option 0 is wrong because HA does not react with additional H⁺; it is the A⁻ (conjugate base) that acts as the proton acceptor.
Question 3 True / False
A buffer solution can resist any amount of added acid or base indefinitely, as long as the solution's pH stays within one unit of the pKa.
TTrue
FFalse
Answer: False
Buffers have finite capacity. When the limiting component (either A⁻ or HA) is consumed by added acid or base, buffering fails and pH changes sharply — just as it would in an unbuffered solution. A buffer containing 0.01 mol each of HA and A⁻ is exhausted by 0.01 mol of added strong acid. The ±1 pH unit range describes the effective operating window under normal conditions, not a guarantee of unlimited resistance. 'Buffer capacity' is a finite quantity that depends on total moles of conjugate pair.
Question 4 True / False
Diluting a buffer solution with water does not significantly change its pH because the Henderson-Hasselbalch equation depends on the ratio [A⁻]/[HA], which remains constant upon dilution.
TTrue
FFalse
Answer: True
Diluting multiplies both [A⁻] and [HA] by the same factor, leaving their ratio — and therefore the log term in Henderson-Hasselbalch — unchanged. pH stays essentially the same. (A very small second-order effect from changing ionic strength exists but is negligible for most purposes.) This is why standard pH buffers used to calibrate electrodes remain reliable even after minor dilution. Buffer capacity, by contrast, is proportionally reduced — fewer moles are available to neutralize added acid or base.
Question 5 Short Answer
Why is buffer capacity maximized when [A⁻] = [HA], and what is the pH at that point?
Think about your answer, then reveal below.
Model answer: When [A⁻] = [HA], the buffer holds equal reserves of both components — equal ability to absorb added acid (converting A⁻ to HA) or added base (converting HA to A⁻). Neither component is limiting, giving the greatest resistance to disturbance in either direction. At this 1:1 ratio, log([A⁻]/[HA]) = log(1) = 0, so pH = pKa exactly.
As the ratio moves away from 1:1 — say to 10:1 or 1:10 — one component becomes the minority. A buffer at [A⁻]/[HA] = 10:1 has ample capacity to absorb added acid (plenty of A⁻) but little capacity to absorb added base (small HA reserve). Maximum capacity in both directions simultaneously requires the 1:1 ratio. This is the practical reason for choosing a weak acid with pKa close to the target pH: it lets you prepare near the 1:1 ratio and operate at maximum capacity.