What concentrations of acetic acid (pKa = 476) and acetate would be required to

Question

What concentrations of acetic acid (pKa = 476) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Note that the concentration and/or pH valve may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid). (A^-)(HA). 2 Use the mole fraction of acetate to calculate the concentration of acetate 3. Calculate the concentration of acetic acid. Step 1: Reanange the Henderson Hasselbalch equation to solve for [A^-]/[HA] The Henderson-Hasselbelch equation is pH = pKa + log ([A^-]/[HA]) Use the Henderson-Hasselbalch equation to solve for [A^-]/[HA] if the solution is at pH 5.0.

Explanation / Answer

So starting with the H-H equation, you can calculate the ratio of [acetate]/[acetic acid]

5 = 4.76 + log [acetate]/[acetic acid]
[acetate]/[acetic acid] = 1.737

OK, so you know that [acetate] + [acetic acid] = 0.15 M
(Whenever the concentration of a buffer is given like that, that means that the total concentration of the weak acid and its conjugate base is that concentration.)

So, you have two equations and two unknowns:
[acetate]/[acetic acid] = 1.727
[acetate] + [acetic acid] = 0.15 M

You should be able to do basic algebra to get to an answer. What I do is rearrange the first one to give:
[acetate] = 1.727[acetic acid]. Then substitute that into the second one to give:
1.727[acetic acid] + [acetic acid] = 0.15
2.727[acetic acid] = 0.15
[acetic acid] = 0.055 M
[acetate] = 0.15 - 0.055 = 0.095 M

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