Question

A \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is slowly added to \(10.0 \mathrm{~mL}\) of a solution that is \(0.20 M\) in \(\mathrm{Ca}^{2+}\) and \(0.30 \mathrm{M}\) in \(\mathrm{Ag}^{+} .\) (a) Which compound will precipitate first: \(\operatorname{CaSO}_{4}\left(K_{s p}=2.4 \times 10^{-5}\right)\) or \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})\) How much \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation?

Short answer

(a) \(\mathrm{CaSO}_{4}\) will precipitate first. (b) Approximately \(83.3 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation of \(\mathrm{CaSO}_{4}\).

Step-by-step solution

Calculate the initial ion product for both compounds

First, we need to calculate the initial ion product [Ca²⁺][SO₄²⁻] and [Ag⁺][SO₄²⁻] without considering the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution. For \(\operatorname{CaSO}_{4}\): \[[Ca^{2+}] = 0.20 M\] \[[SO_{4}^{2-}] = 0\] So the initial ion product for \(\operatorname{CaSO}_{4}\) is: \[((0.20)(0))=0\] For \(\mathrm{Ag_{2}SO}_{4}\): \[[Ag^{+}] = 0.30 M\] \[[SO_{4}^{2-}] = 0\] So the initial ion product for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) is: \[((0.30)(0))=0\]

Calculate the Q value as the solution is added

When the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is added, the concentration of \(SO_{4}^{2-}\) ions increases. Let the volume of added solution be \(V\). For \(\operatorname{CaSO}_{4}\): \[[SO_{4}^{2-}] = (1.0M \cdot V)/(V+10)\] So the ion product (Q) after adding the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is: \[Q_{CaSO_4} = [Ca^{2+}][SO_{4}^{2-}] = (0.20)((1.0M \cdot V)/(V+10))\] For \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\): \[[SO_{4}^{2-}] = (1.0M \cdot V)/(V+10)\] So the ion product (Q) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\), after addition of the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is: \[Q_{Ag_2SO_4} = [Ag^{+}][SO_{4}^{2-}] = (0.30)((1.0M \cdot V)/(V+10))\]

Determine the first precipitate

Whichever compound precipitates first will have its Q value reach the \(K_{sp}\) first. Since both Q expressions have the same \([SO_{4}^{2-}]\) term, we can determine which reaches their \(K_{sp}\) first by comparing the ratio of Q/Ksp for each compound. Solve for "V" to determine which compound has a smaller volume required to start precipitation. For \(\operatorname{CaSO}_{4}\): \[Q_{CaSO_4}/K_{sp}=(0.20)((1.0\cdot V)/(V+10))/(2.4 \times 10^{-5})=1\] For \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\): \[Q_{Ag_2SO_4}/K_{sp}=(0.30)((1.0\cdot V)/(V+10))/(1.5 \times 10^{-5})=1\] Solve for V for both equations.

Calculate the Volumes

For \(\operatorname{CaSO}_{4}\): \[V_{CaSO_4} = \frac{0.20 \cdot 10 + 2.4 \times 10^{-5} \cdot V}{2.4 \times 10^{-5}} - 10\] \[V_{CaSO_4} \approx 83.3 mL\] For \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\): \[V_{Ag_2SO_4} = \frac{0.30 \cdot 10 + 1.5 \times 10^{-5} \cdot V}{1.5 \times 10^{-5}} - 10\] \[V_{Ag_2SO_4} \approx 199.3 mL\] Since \(V_{CaSO_4} < V_{Ag_2SO_4}\), \(\mathrm{CaSO}_{4}\) will precipitate first.

Answer the question

(a) The compound that precipitates first is \(\mathrm{CaSO}_{4}\). (b) To initiate the precipitation of \(\mathrm{CaSO}_{4}\), approximately \(83.3 mL\) of \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or \(K_{sp}\), is a crucial concept. It measures the solubility of a compound in water. This constant is specific to each compound at a given temperature.When representing \(K_{sp}\), we look at the concentrations of the ions in a saturated solution. For example, in the case of \(CaSO_{4}\), \(K_{sp}\) would be expressed as:\[K_{sp} = [Ca^{2+}][SO_{4}^{2-}]\]The idea behind \(K_{sp}\) is simple:

• If the ion product of the solution is less than \(K_{sp}\), the solution is unsaturated, and no precipitate forms.
• If the ion product equals \(K_{sp}\), the solution is saturated and on the verge of forming a precipitate.
• If the ion product exceeds \(K_{sp}\), the solution is supersaturated, and a precipitate will form.

Understanding \(K_{sp}\) helps predict if a reaction will form a precipitate, which is a solid that separates from the liquid solution. By knowing the \(K_{sp}\), chemists can anticipate the behavior of different compounds when mixed.

Ion Product

The ion product, denoted as \(Q\), represents the product of the concentrations of the ions present in a solution at any given time. It's similar to the solubility product constant \(K_{sp}\), but unlike \(K_{sp}\), it's calculated under non-equilibrium conditions.For instance, in determining whether \(CaSO_{4}\) or \(Ag_2SO_{4}\) will precipitate first, we need to calculate the ion product for each.
This is done as the sulfate ions \(([SO_4^{2-}])\)from \(Na_2SO_4\) are added to the solution:

• The ion product for \(CaSO_4\) is \(Q = [Ca^{2+}][SO_{4}^{2-}]\).
• The ion product for \(Ag_2SO_4\) is \(Q = [Ag^{+}]^2[SO_{4}^{2-}]\).

When the ion product \(Q\) of a compound in a solution exceeds its \(K_{sp}\), a precipitate begins to form. Comparing \(Q\) to \(K_{sp}\) allows chemists to tell whether conditions are just right for a solid to formor whether more ions must dissolve before precipitation occurs.

Precipitation Reaction

When a precipitation reaction occurs, a solid is formed from two substances in a solution. This process depends on the interaction between ions and often involves a shift from a dissolved to a solid state.In our example, as \(Na_2SO_4\) is added to a solution containing \(Ca^{2+}\) and \(Ag^{+}\), we are interested in which will form a precipitate first.Precipitation occurs when:

• The concentration of ions surpasses the \(K_{sp}\), resulting in a shift from dissolved ions back to a solid.
• Conditions favor the formation of a stable solid structure.

In chemical terms,\(CaSO_4\) forms a precipitate first when enough \(SO_4^{2-}\) ions have been added to satisfy its ion product reaching > \(K_{sp}\). Precipitates generally form because the solid state is more stable than the high concentration of dissolved ions. This stability drives the reaction, resulting in a solid formation.