To understand how buffers use reserves of conjugate acid and conjugate base to c
ID: 484193 • Letter: T
Question
To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH. A buffer is a mixture of a conjugate acid-base pair. In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid, CH_3COOH, and its conjugate base, the acetate ion CH_3COO^-. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g., sodium acetate NaCH_3COO). Buffers work because the conjugate acid-base pair work together to neutralize the addition of H^+ or OH^- ions. Thus, for example, if H^+ ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H^+ with the conjugate base: H^+ +CH_3COO^- rightarrow CH_3COOH Similarly, any added OH^- ions will be neutralized by a reaction with the conjugate acid: OH^- +CH_3COOH rightarrow CH_3COO^- + H_2 O This buffer system is described by the Henderson-Hasselbalch equation pH = pK_a + log [conjugate base]/[conjugate acid] A beaker with 2.00 times 10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.30 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pK_a of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.Explanation / Answer
Given that pH of the buffer = 5.0
pKa = 4.74
[acetate ion/CH3COO-] = molarity x volume in Litres = 0.1 M x 0.2 L = 0.02 mol
[HCl] = molarity x volume in Litres = 0.3 M x 0.0043 L = 0.00129 mol
CH3COO- + HCl ------------> CH3COOH + Cl-
0.02 mol 0.00129 mol 0
----------------------------------------------------------------------------------------------
0.02-0.00129 0 0.00129 mol
= 0.01871 mol
Therefore,
[CH3COOH] = 0.00129 mol
[CH3COO-] = 0.01871 mol
Henderson-Hasselbalch equation is
pH = pKa + log [CH3COO-]/[CH3COOH]
= 4.74 + log (0.01871 mol/0.00129 mol)
= 3.58
pH = 3.58
Therefore,
change in pH , pH = 3.58 - 5.0 = -1.42
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