Question

Phosphoric acid is a triprotic acid (Ka1 = 6.9× 10–3, Ka2 = 6.2× 10–8, and Ka3 = 4.8× 10–13). To find the pH of a buffer composed of H2PO4–(aq) and HPO42–(aq), which pKa value would you use in the Henderson-Hasselbalch equation?

Calculate the pH of a buffer solution obtained by dissolving 15.0 g of KH2PO4(s) and 33.0 g of Na2HPO4(s) in water and then diluting to 1.00 L.

Answer 1

Before determining "which pK_a value would you use in the Henderson-Hasselbalch equation?", we first of all need to know "What is Henderson-Hasselbalch equation?" briefly.

So, this equation is mainly used to determine the pH value of a Buffer solution using the acidity or pK_a value of the involved acid-base equilibrium. (It can also be determined with pK_b value, but that's not our cocern right now).

Now, Phosphoric acid being a triprotic acid dissociates in the following way ---

H₃PO₄    H₂PO₄^- + H⁺ ; K_a1 = 6.9 10^-3

H₂PO₄^- HPO₄^2- + H⁺ ; K_a2 = 6.2 10^-8

HPO₄^2-    PO₄^3- + H⁺ ; K_a3 = 4.8 10^-13

Here in the question the concerned buffer is composed of H₂PO₄^- (aq) and HPO₄^2- (aq). So, the equilibrium of our interest is the 2nd one and the pK_a value that should be used in the Henderson-Hasselbalch equation is pK_a2 (= -log(6.2 10^-8) = 7.2076 ).

Now the given problem also deals with the same H₂PO₄^- (aq) and HPO₄^2- (aq) equilibrium and to calculate the pH of this given buffer solution we need to apply Henderson-Hasselbalch equation.

Where, = Molar concentration of the involved acid/acid-salt, here, HA = H₂PO₄^-

= Molar concentration of the Conjugate Base, here, A^- = HPO₄^2-

and pK_a= pK_a2 = 7.2076

Now as the solution is diluted to 1 litre, its molar concentration = the number of moles present in the solution. So, = 15/136.1 = 0.1102 mol. L^-1 (M.W. of KH₂PO₄ = 136.1)

and = 33/142 = 0.2324 mol. L^-1 (M.W. of Na₂HPO₄ = 142)

So, putting all the values in the Henderson-Hasselbalch equation we get---

pH = 7.2076 + log₁₀(0,2324/0.1102)

    pH = 7.5317