Estimate the pH of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (Na₂HPO₄) and 1.25 mole of hydrochloric acid. The pK_a values for phosphoric acid are 2.1, 7.2, 12.4.

Question

Quizplus · Accepted Answer

The balanced equation for the reaction between disodium phosphate and hydrochloric acid is:

Na2HPO4 + 2 HCl → 2 NaCl + H3PO4

From the balanced equation, we can see that the acid produced by the reaction is phosphoric acid (H3PO4). Phosphoric acid is a triprotic acid, which means that it can donate three protons (H+) per molecule.

The reaction between 1.0 mole of disodium phosphate and 1.25 mole of hydrochloric acid will fully consume the hydrochloric acid (since hydrochloric acid is the limiting reagent). The number of moles of phosphoric acid produced can be calculated as follows:

1.25 mole HCl x (1 mole H3PO4 / 2 mole HCl) = 0.625 mole H3PO4

Now we need to determine the pH of a solution containing 0.625 mole H3PO4. To do this, we need to consider each step of the ionization of phosphoric acid.

Step 1: H3PO4 ⇌ H+ + H2PO4-
pKa1 = 2.1

At equilibrium, we have:

[H+] [H2PO4-] / [H3PO4] = 10^-pKa1

[H+] [H2PO4-] / 0.625 = 10^-2.1

For simplicity, we can assume that x is much smaller than 0.625. This allows us to make the approximation:

[H+] [H2PO4-] / 0.625 ≈ [H+]^2 / 0.625

[H+]^2 / 0.625 = 7.94 x 10^-3

[H+] = √(7.94 x 10^-3 x 0.625) = 0.049 M

pH = -log[H+] = 1.31

Step 2: H2PO4- ⇌ H+ + HPO42-
pKa2 = 7.2

At equilibrium, we have:

[H+] [HPO42-] / [H2PO4-] = 10^-pKa2

[H+] [HPO42-] / (0.625 - [H+]) = 10^-7.2

Again, we can make the approximation that x is much smaller than 0.625 - [H+]. This simplifies the equation:

[H+] [HPO42-] / 0.625 ≈ [H+]^2 / 0.625

[H+]^2 / 0.625 = 5.01 x 10^-8

[H+] = √(5.01 x 10^-8 x 0.625) = 2.23 x 10^-4 M

pH = -log[H+] = 3.65

Step 3: HPO42- ⇌ H+ + PO43-
pKa3 = 12.4

At equilibrium, we have:

[H+] [PO43-] / [HPO42-] = 10^-pKa3

[H+] [PO43-] / ([H2PO4-] - [H+]) = 10^-12.4

Since [H+] is much smaller than [H2PO4-], we can approximate [H2PO4-] - [H+] as 0.625:

[H+] [PO43-] / 0.625 ≈ [H+]^2 / 0.625

[H+]^2 / 0.625 = 3.98 x 10^-14

[H+] = √(3.98 x 10^-14 x 0.625) = 5.00 x 10^-8 M

pH = -log[H+] = 7.30

The pH of the resulting solution is in the range of 2.1 < pH < 7.2, which corresponds to choice C.

Estimate the PH of the Resulting Solution Prepared by Mixing