A weak acid (H3PO4) and its conjugate base (NaH2PO4) will make a buffer solution.
H3PO4 = H+ + H2PO4-, Ka1 = 7.1x10^-3 (I)
H2PO4- = H+ + (HPO4)2-, Ka2 = 6.3x10^-8 (II)
(HPO4)2- = H+ + (PO4)3-, Ka3 = 4.5x10^-13 (III)
Assumption: Since Ka3 << Ka2 << Ka1, the [H+] produced from (II) and (III) is negligible compared to that from (I).
(a) 100 mL solution A + 100 mL solution B:
Initial concentrations: [H3PO4] = 0.05 M; [H2PO4-] = 0.05 M.
Reaction: H3PO4 = H+ + H2PO4-
Intial: 0.05 0 0.05
React: x x x
Final: (0.05-x) x (0.05+x)
At equilibrium, Ka1 = [H+]*[H2PO4-]/[H3PO4] = x*(0.05+x) / (0.05-x) = 7.1x10^-3
[H+] = x = 5.7x10^-3 M
pH = -log[H+] = 2.2
(Notes: If we assume that (0.05+x) ~ 0.05 ~ (0.05-x), the math would be more simple and [H+] = 7.1x10^-3 M or pH = 2.1.)
Check the previous assumption:
Given that [H+] = 5.7x10^-3 M and [H2PO4-] = 4.43x10^-2 M at equilibrium, let's consider reaction (II)
H2PO4- = H+ + (HPO4)2-, Ka2 = 6.3x10^-8
Ka2 = [H+][(HPO4)2-]/[H2PO4-]
--> [(HPO4)2-] = 4.9x10^-7 M << 5.7x10^-3 M. That means the number of protons produced from (II) is insignificant, and the above assumptioin is valid.
(b) 100 mL solution A' (0.01 M H3PO4) + 100 mL solution B' (0.01 M NaH2PO4): In other words, we dilute the buffer solution in (a) 10 times.
We can repeat the exercise and come up with new values for [H+] and pH, but I bet the pH would remain about the same (~ 2.2). This is just a characteristic of a buffer solution. The pH of a buffer solution does not change much as it is diluted or as a small amount of acid or base is added.
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