What concentrations of acetic acid (pka = 4.76) and acetate would be required to prepare a 0.15 m buffer solution at ph 5.0? note that the concentration and/or ph value 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: rearrange the henderson hasselbalch equation to solve for (a-)/(ha) if the solution is at ph 5.0

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 =  [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!