H3PO4 ⇄ H+ + H2PO4− Ka1 = 7.2 × 10^−3 H2PO4− ⇄ H+ + HPO4^2− Ka2 = 6.3 × 10^−8 HPO4^2− ⇄ H+ + PO4^3− — Acids and Bases Chemistry Question

STEPS:

1. Understand the dissociation of the starting salts: When the two highly soluble sodium salts are dissolved in water, they dissociate completely into their constituent ions:
* $\text{NaH}_2\text{PO}_4(s) \rightarrow \text{Na}^+(aq) + \text{H}_2\text{PO}_4^-(aq)$
* $\text{Na}_2\text{HPO}_4(s) \rightarrow 2\ \text{Na}^+(aq) + \text{HPO}_4^{2-}(aq)$
2. Calculate the concentrations of the major species in the mixture:
* Mixing equal volumes ($50\text{ mL}$ each) of both $1\text{ M}$ solutions doubles the total volume to $100\text{ mL}$, which halves the individual species concentrations:
* $[\text{H}_2\text{PO}_4^-] = 1\text{ M} \times \frac{50\text{ mL}}{100\text{ mL}} = \mathbf{0.5\text{ M}}$
* $[\text{HPO}_4^{2-}] = 1\text{ M} \times \frac{50\text{ mL}}{100\text{ mL}} = \mathbf{0.5\text{ M}}$
* The total sodium ion concentration is even higher because it is released by both salts:
* $[\text{Na}^+] = \frac{(1 \times 50\text{ mmol}) + (2 \times 50\text{ mmol})}{100\text{ mL}} = \mathbf{1.5\text{ M}}$
3. Analyze the resulting buffer equilibrium: Because $\text{H}_2\text{PO}_4^-$ (a weak acid) and $\text{HPO}_4^{2-}$ (its conjugate base) are present in high, equal concentrations ($0.5\text{ M}$), they establish a classic buffer system governed by the second acid-dissociation constant:
$$\text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \quad K_{a2} = 6.3 \times 10^{-8}$$
Because the concentrations of $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$ are equal, the hydronium concentration $[\text{H}^+]$ is approximately equal to $K_{a2}$ ($6.3 \times 10^{-8}\text{ M}$), yielding a $\text{pH} \approx 7.2$.
4. Evaluate the third ionization step to find $[\text{PO}_4^{3-}]$: The hydrogen phosphate ion ($\text{HPO}_4^{2-}$) can dissociate further to form the phosphate ion ($\text{PO}_4^{3-}$):
$$\text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \quad K_{a3} = 4.5 \times 10^{-13}$$
Using the equilibrium expression for $K_{a3}$:
$$K_{a3} = \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]}$$
Rearranging to solve for the concentration of phosphate:
$$[\text{PO}_4^{3-}] = K_{a3} \times \frac{[\text{HPO}_4^{2-}]}{[\text{H}^+]} \approx (4.5 \times 10^{-13}) \times \frac{0.5\text{ M}}{6.3 \times 10^{-8}\text{ M}} \approx \mathbf{3.6 \times 10^{-6}\text{ M}}$$
5. Compare the magnitudes of all species:
* $[\text{Na}^+] = 1.5\text{ M}$
* $[\text{H}_2\text{PO}_4^-] \approx 0.5\text{ M}$
* $[\text{HPO}_4^{2-}] \approx 0.5\text{ M}$
* $[\text{PO}_4^{3-}] \approx 3.6 \times 10^{-6}\text{ M}$
Clearly, the concentration of $\text{PO}_4^{3-}$ is many orders of magnitude smaller than any of the other species, making Option D the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: $\text{Na}^+$ is a spectator ion that does not react with water or participate in any acid-base equilibria. Since both salts contain sodium, it is present in the highest concentration in the solution ($1.5\text{ M}$).
• Option B is incorrect: $\text{HPO}_4^{2-}$ is one of the primary components used to make the buffer. Because its acid dissociation and base hydrolysis are extremely weak (governed by very small equilibrium constants), its concentration at equilibrium remains very close to its starting concentration of $0.5\text{ M}$.
• Option C is incorrect: $\text{H}_2\text{PO}_4^-$ is the other major component of the buffer. Like $\text{HPO}_4^{2-}$, its dissociation is minimal, so its equilibrium concentration is also extremely high ($\approx 0.5\text{ M}$).