Use the HH equation. The way I read the problem you have a solution that is 0.2M in propionic acid and another solution 0.2M in sodium propionate.
pKa = -log Ka = about 4.89 but you should confirm that and adjust the figure to your liking.
4.65 = 4.89 + log (base/acid)
b/a = 0.575
Then base = 0.575*acid
Let x = mL acid and 50-x = mL base.
millimols acid = 0.2M*xmL acid
millimols base = 0.2M*(50-xmL acid) so
0.2*(50-xmLacid) = 0.575*0.2M*xmL acid
10-0.2xmL acid = 0.115x*mL acid
10 = 0.315x
x = 10/0.315 = 31.75 mL acid
50-x = 18.25 mL base
I always check these things to make sure they come out to the desired pH.
M base = 18.25*0.2/50 = 0.073
M acid = 31.75*0.2/50 = 0.127
pH = 4.89 + log(0.073/0.127) = 4.65. voila!
Note: I've carried this to too many significant figures and they should be adjusted to only two places so 32 and 18 would be appropriate I think.
Propionic acid has a ka of 1.3x10^(-5), 50.0ml of a buffer solution having a pH of 4.65 is to be prepared using a 0.20 M solutions of propionic acid and sodium propionate. How many ml of each of these solutions must be mixed?
Thanks in Advanced !
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