Questions 22-24 refer to the following information. A group of students was asked to recover Cu(s) f — Acids and Bases Chemistry Question

STEPS:

1. Analyze the chemical environment at the end of step 2:
* In Step 2, the solid copper phosphate ($\text{Cu}_3(\text{PO}_4)_2$) precipitate is dissolved in a highly concentrated strong acid, $3.0\text{ M }\text{HCl}(aq)$.
* This dissolution reaction frees the copper and phosphate ions into solution, where the phosphate protonates to form phosphoric acid ($\text{H}_3\text{PO}_4$):
$$\text{Cu}_3(\text{PO}_4)_2(s) + 6\ \text{H}^+(aq) \rightarrow 3\ \text{Cu}^{2+}(aq) + 2\ \text{H}_3\text{PO}_4(aq)$$
* We are told that the resulting solution has a pH of approximately 1. Since $\text{pH} = -\log[\text{H}^+]$, a pH of 1 means:
$$[\text{H}^+] \approx 10^{-1}\text{ M} = 0.1\text{ M}$$
This high concentration of hydrogen ions is primarily due to the excess hydrochloric acid used to dissolve the precipitate.

2. Analyze the successive ionization steps of $\text{H}_3\text{PO}_4$:
Phosphoric acid is a weak triprotic acid that ionizes in three sequential, increasingly difficult steps:
* First ionization: $\text{H}_3\text{PO}_4(aq) \rightleftharpoons \text{H}^+(aq) + \text{H}_2\text{PO}_4^-(aq) \quad K_{a1} = 7.5 \times 10^{-3}$
* Second ionization: $\text{H}_2\text{PO}_4^-(aq) \rightleftharpoons \text{H}^+(aq) + \text{HPO}_4^{2-}(aq) \quad K_{a2} = 6.2 \times 10^{-8}$
* Third ionization: $\text{HPO}_4^{2-}(aq) \rightleftharpoons \text{H}^+(aq) + \text{PO}_4^{3-}(aq) \quad K_{a3} = 4.2 \times 10^{-13}$

3. Compare relative concentrations using the equilibrium constants:
To determine which conjugate base is present in the lowest concentration, we look at how the ratio of ionized species behaves under a high $[\text{H}^+]$ concentration of $0.1\text{ M}$:
* For the first step:
$$\frac{[\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} = \frac{K_{a1}}{[\text{H}^+]} \approx \frac{7.5 \times 10^{-3}}{0.1} = 0.075$$
This indicates that only a small fraction of the $\text{H}_3\text{PO}_4$ has ionized to $\text{H}_2\text{PO}_4^-$.
* For the second step:
$$\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = \frac{K_{a2}}{[\text{H}^+]} \approx \frac{6.2 \times 10^{-8}}{0.1} = 6.2 \times 10^{-7}$$
The concentration of $\text{HPO}_4^{2-}$ is less than one-millionth of the $\text{H}_2\text{PO}_4^-$ concentration.
* For the third step:
$$\frac{[\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]} = \frac{K_{a3}}{[\text{H}^+]} \approx \frac{4.2 \times 10^{-13}}{0.1} = 4.2 \times 10^{-12}$$
The concentration of $\text{PO}_4^{3-}$ is twelve orders of magnitude smaller than the already minute concentration of $\text{HPO}_4^{2-}$.

4. Establish the final concentration hierarchy:
Combining these ratios yields the following relative abundance for the phosphorus species in the highly acidic solution:
$$[\text{H}_3\text{PO}_4] \gg [\text{H}_2\text{PO}_4^-] \gg [\text{HPO}_4^{2-}] \gg [\text{PO}_4^{3-}]$$
Because each successive deprotonation is heavily suppressed by the high concentration of hydronium ions (a stark manifestation of the common-ion effect on polyprotic systems), the fully deprotonated phosphate ion, $\text{PO}_4^{3-}$, has the absolute lowest concentration among the species listed. This corresponds to Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Because the solution is highly acidic ($\text{pH} \approx 1$), $\text{H}^+$ is one of the most abundant species in the entire mixture, with a high concentration of approximately $0.1\text{ M}$.
• Option B is incorrect: Dihydrogen phosphate ($\text{H}_2\text{PO}_4^-$) represents the product of the first ionization step. Because $K_{a1} = 7.5 \times 10^{-3}$ is the largest dissociation constant by far, its concentration is significantly higher than those of the subsequent ionized species.
• Option C is incorrect: While the hydrogen phosphate ion ($\text{HPO}_4^{2-}$) is present in an extremely low concentration due to the small value of $K_{a2} = 6.2 \times 10^{-8}$, it is still roughly eleven orders of magnitude *more* abundant than the fully deprotonated phosphate ion ($\text{PO}_4^{3-}$).