Buffers must be prepared from a weak acid and its weak conjugate base or a weak base and its weak conjugate acid.
In this example, HCO3^1- is the weak acid (HA), and CO3^2- is the weak conjugate base (A-).
Whenever you have a buffer solution, think Henderson Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Since molarity/molarity is simply a ratio, you can use moles/moles or any units because units cancel.
This example gave the ratio of acid to base...we must invert it to give us the ratio of base to acid (4 inverted is 0.25)
pKa = - log Ka = - log (4.7 x 10^-11) = 10.33
A) pH = 10.33 + log (0.25)
pH = 10.33 - 0.602 = 9.73
B) If we add HCl (a strong acid), the weak base (CO3^2- or A-) will neutralize it and will be converted to HCO3^1-...A- will go down by .01 moles and the HA will go up by .01 moles.
Since the original ratio of base to acid was 1 to 4, the new ratio is now: 1 - .01 to 4 + .01. We can assume the original was 1 mole to 4 moles (it is a valid assumption to get approximate pH values if they only give you a ratio).
pH = 10.33 + log (.99/4.01)
pH = 10.33 + log (.247)
pH = 10.33 - 0.6075
pH = 9.72
By adding an acid, the pH went down...so this makes sense.
C) If we add NaOH (a strong base), the weak acid (HCO3^1- or HA) will neutralize it and will be converted to CO3^2-...HA will go down by .02 moles and A- will go up by .02 moles.
Since the original ratio of base to acid was 1 to 4, the new ratio is now: 1 + .02 to 4 - .02. We can assume the original was 1 mole to 4 moles.
pH = 10.33 + log (1.02/3.98)
pH = 10.33 + log (.256)
pH = 10.33 - .5913
pH = 9.74
By adding a base, the pH went up...so this makes sense.
Note: very small change in over all pH occurred by adding acid or base. That is the power of buffered solutions!
