Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer with a total volume of 200. mL. Assume all stock solutions of the conjugate pair are 1.00 M. Show all of your work.

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

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3

explanation: