Ethanoate Buffer Under Fermentation Acid Load
A student prepares a buffer for a fermentation experiment by mixing 50.0 mL of 0.100 mol L−1 ethanoic acid (Ka = 1.74 × 10−5) with 50.0 mL of 0.100 mol L−1 sodium ethanoate. The pH meter reads 4.76. The student then adds 5.00 mL of 0.100 mol L−1 HCl and records the new pH. A supervisor says the buffer is 'used up' and should be replaced, but the student argues the pH change is small. The fermentation must stay between pH 4.5 and 5.0 for 48 hours. The expected acid production is 0.0040 mol of H+ per litre of buffer. The student wants to know whether the buffer will last and whether the supervisor's claim is justified.
1. Calculate the pH before and after adding the 5.00 mL of 0.100 mol L−1 HCl. Show the mole amounts and the ratio used.
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Initial moles: n(HA) = 0.0500 L × 0.100 mol L−1 = 5.00 × 10−3 mol; n(A−) = same = 5.00 × 10−3 mol. Initial pH = pKa + log([A−]/[HA]) = 4.76 + log(1.00) = 4.76.
Added HCl: n(H+) = 0.00500 L × 0.100 = 5.00 × 10−4 mol. This converts A− to HA: n(HA) = 5.00 × 10−3 + 5.00 × 10−4 = 5.50 × 10−3 mol; n(A−) = 5.00 × 10−3 − 5.00 × 10−4 = 4.50 × 10−3 mol. Total volume = 0.105 L but cancels in the ratio. pH = 4.76 + log(4.50/5.50) = 4.76 + log(0.818) = 4.76 − 0.087 = 4.67. The pH falls by only 0.09 units, so the buffer is responding, not simply 'used up'.
2. Calculate how many moles of H+ the 105 mL buffer can absorb before the pH reaches 4.5. Compare this with the expected acid production and decide whether the buffer should last 48 hours.
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At pH 4.5, [A−]/[HA] = 10^(4.5−4.76) = 10^(−0.26) = 0.55. Let x = moles of H+ absorbed. Then n(HA) = 5.00 × 10−3 + x and n(A−) = 5.00 × 10−3 − x. (5.00 × 10−3 − x)/(5.00 × 10−3 + x) = 0.55. 5.00 × 10−3 − x = 2.75 × 10−3 + 0.55x 2.25 × 10−3 = 1.55x x = 1.45 × 10−3 mol H+ in 0.105 L. Expected H+ in 0.105 L = 0.0040 mol L−1 × 0.105 L = 4.2 × 10−4 mol. This is less than 1.45 × 10−3 mol, so on this estimate the buffer has enough capacity to keep pH above 4.5 for the expected acid load. However, if acid production is faster or other acids are present, the buffer could still fail; the calculation supports 'will last', not a guarantee.
3. Explain why adding 5.00 mL of water instead of HCl changes the pH only slightly, while adding the same volume of HCl changes it measurably.
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pH depends on the ratio [A−]/[HA] through pH = pKa + log([A−]/[HA]). Adding water dilutes both HA and A− by the same factor, so the ratio, and therefore the pH, is almost unchanged (apart from small activity effects). Adding HCl does not simply dilute the buffer: H+ reacts with A− to form HA, so the numerator decreases and the denominator increases. That changes the ratio and lowers the pH. The buffer resists pH change by converting added H+ into weak acid, but it cannot prevent the ratio from changing once A− is consumed.