This problem is a little confusing because of the wording; i.e., it isn't easy to tell when we are talking about system 2 and when we aren't.
The best buffer to use is one that has pKa = pH you want for the solution. So convert Ka for each system to pKa. I get something like
for 4E-3 = 2.4
for 5E-4 = 3.3
for 6E-5 = 4.2
Take your pick from my initial remarks.
b. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
Set pH = 3.50 and solve for base/acid ratio.
c. System 2 is
pH = 3.30 + log (base/acid)
Adding 0.03 moles HCl to the buffer. The buffer, to resist a change in pH, will react H^+ with the A^- base, as
A^- + H^+ ==> HA
Therefore, addition of 0.03 mole HCl will decrease A^- by 0.03 and increase HA by 0.03. Make those change in log(base/acid) and calculate new pH.
d. Set pH in the H-H equation to 3.5 + 0.7 = ??(I suppose this is what is meant--the problem doesn't say 0.7 from what initial value) and calculate b/a ratio. That will tell you what the new HA and A must be and you add NaOH to accomplish that.
You have to design a buffer based on one of the systems below
System 1: HA1 and A1-1 K= 4 E-3
System 2: HA2 and A2-1 K= 5 E-4
System 3: HA3 and A3-1 K= 6 E-5
For the following parts assume that the buffer above in system2 has been constructed with [HA] =1M and that there is no change in volume when acids or bases are added
a) Pick the best system to make a buffer with a pH of 3.5.
b) For which ever system you have chosen find the [A^-1]/[HA] required to make the pH equal to 3.5
c) Find the new pH if .03 moles of HCl is added to 100 ml of buffer in system 2
d) Find the maximum mass of NaOH which could be added to 100 ml of the original buffer without changing the pH by more than .7 pH units
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