To save some space and a lot of typing, let's call CH3COOH = acid = a and likewise call CH3COONa = base = b. The the Henderson-Hasselbalch equation (HH equation) is
pH = pKa + log (CH3COONa/CH3COOH) = pKa + log b/a.The pKa for acetic acid is 4.76.
For this buffer to have a buffer capacity of 1 mol for a +/- 1 pH, we want
3.76 = 4.76 + log b/a
-1 = log b/a
b/a = 0.1 = (CH3COONa/CH3COOH)
When strong acid, for example HCl, is added you have this reaction.
CH3COONa + H^+ ==> CH3COOH. Let x = moles CH3COONa = moles CH3COOH so that
I..........x...............0.................x
add......................1...........................
C.........-1..............-1............+1
E.........x-1.............0...............x+1
Then that ratio of b/a looks like this.
(x-1)/(x+1) = 0.1
Solve for x. That looks like 1.22 to me; therefore, (CH3COOH) = (CH3COONa) = 1.22 mols for 1 L
solution. Therefore, you want to make up a 1.22 M solution in CH3COOH and 1.22 M in CH3COONa.
That will be 60 x 1.22 = 73.2 g CH3COOH and
82 x 1.22 = 100 g CH3COONa in 1 L of solution. That will have a pH of 4.76 (theoretical anyway) with a buffer capacity of 1 mole for +/- 1 pH.
I like to check these things this way.
CH3COONa + H^+ ==> CH3COOH + Na^+
I.....1.22............0................1.22
add..................1....................
C.......-1...........-1..................+1
E.......0.22..........0................2.22
Then plug this into the HH equation:
pH = 4.76 + log (0.22/2.22)
pH - 4.76 -1
pH = 3.76
voila!
You wish to prepare a pH 4.76 acetic acid-sodium acetate buffer with a buffer capacity of 1M per ph. What concentration of acetic acid and sodium acetate are needed
1 answer