Question

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}\). (a) What con- centration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}:\) \(\left.K_{s p}=3.2 \times 10^{-7} .\right)\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}\) when the second cation begins to precipitate?

Short answer

(a) A concentration of \(1.1 \times 10^{-8} M\) sulfate ions is necessary for precipitation to begin. (b) The cation that precipitates first is Ba^2+. (c) The concentration of sulfate ions when Sr^2+ begins to precipitate is \(3.2 \times 10^{-5} M\).

Step-by-step solution

Write down the solubility product expressions

For the precipitation reactions, the solubility product expressions are: \(K_{sp,\text{BaSO4}} = [Ba^{2+}][SO_4^{2-}]\) and \(K_{sp,\text{SrSO4}} = [Sr^{2+}][SO_4^{2-}]\) Where \(K_{sp,\text{BaSO4}} = 1.1 \times 10^{-10}\) and \(K_{sp,\text{SrSO4}} = 3.2 \times 10^{-7}\), and the concentrations of the ions are given in moles per liter.

Determine sulfate ion concentration for precipitation

We need to find the concentration of sulfate ions needed for precipitation to occur. We will use the solubility product expressions to do so. For BaSO4: \([SO_4^{2-}] = \frac{K_{sp,\text{BaSO4}}}{[Ba^{2+}]}\) \([SO_4^{2-}] = \frac{1.1 \times 10^{-10}}{0.010}\) For SrSO4: \([SO_4^{2-}] = \frac{K_{sp,\text{SrSO4}}}{[Sr^{2+}]}\) \([SO_4^{2-}] = \frac{3.2 \times 10^{-7}}{0.010}\) Calculate the concentrations.

Compare the calculated concentrations

Calculate the concentrations in Step 2: For BaSO4: \([SO_4^{2-}] = 1.1 \times 10^{-8} M\) For SrSO4: \([SO_4^{2-}] = 3.2 \times 10^{-5} M\) Compare these concentrations to find out which cation precipitates first and the necessary concentration of sulfate ions for precipitation to begin. The cation with the lower sulfate concentration will precipitate first.

Determine the first precipitating cation

Comparing the calculated sulfate concentrations, we find that the concentration of sulfate ions needed for BaSO4 precipitation (1.1 x \(10^{-8} M\)) is lower than that required for SrSO4 precipitation (3.2 x \(10^{-5} M\)). Hence, the answer to (a) is that a concentration of 1.1 x \(10^{-8} M\) sulfate ions is necessary for precipitation to begin. And the answer to (b) is that the cation that precipitates first is Ba^2+.

Find the concentration of sulfate ions for the second precipitation

We have already determined that the second cation to precipitate is Sr^2+. To find the concentration of sulfate ions when this occurs, we can use the calculated concentration from Step 2 for SrSO4 precipitation. The answer to (c) is that the concentration of sulfate ions when Sr^2+ begins to precipitate is 3.2 x \(10^{-5} M\).

Key concepts

Precipitation Reactions

When two solutions that contain soluble salts are mixed, sometimes an insoluble solid, known as a precipitate, forms. This process is called a precipitation reaction. In the context of the provided exercise, such a reaction occurs when a solution of sodium sulfate (Na₂SO₄) is slowly added to a mixed solution containing barium ((Ba²⁺) and strontium (Sr²⁺) cations. To understand when precipitation begins, one must calculate the ion concentration at which the solution becomes supersaturated with the precipitating ion.

The start of precipitation is indicated by the solubility product constant, K_sp, specific to the given ionic compound. It is a unique value representing the maximum product of ion concentrations that can exist in a solution before precipitation occurs. If the actual product of ion concentrations in solution exceeds the solubility product, the excess ions will form a precipitate. This is an important concept because it allows us to calculate exactly when the precipitate will form, thus giving insight into the stoichiometry and dynamics of the reaction.

Solubility Equilibrium

Solubility equilibrium is the point at which a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The solubility product (K_sp) represents the constant for this equilibrium. The key thing to understand is that solubility equilibrium is all about balance. If nothing changes in the system, the levels of dissolved ions and the undissolved solid will remain constant. However, if more ions are introduced or if conditions change, the equilibrium may shift.

Detailed calculations of the concentration of sulfate ions (SO₄^2-) necessary for the precipitation of barium sulfate (BaSO₄) and strontium sulfate (SrSO₄) help us understand how the equilibrium changes when we alter the ion concentration in the solution. By performing these calculations, we can predict which compound will precipitate first; thus, understanding this equilibrium concept is essential in practical applications, such as in waste treatment or in the separation of ions in analytical chemistry.

Ion Concentration

Ion concentration is a critical factor in determining the solubility and precipitation of substances in a solution. It is defined as the amount of an ion in a solution, usually expressed in moles per liter (M). In our exercise, we are dealing with the concentration of sulfate ions (SO₄^2-) in a solution that already contains barium (Ba²⁺) and strontium (Sr²⁺) cations at known concentrations.

Calculations to determine at what point additional sulfate will result in the precipitation of BaSO₄ and SrSO₄ are directly related to the concept of ion concentration. The concentration of ions leads to an understanding of the concentrations required for the solution to reach a point where precipitation starts, known as the molar solubility. In practical situations, knowing the ion concentration can help in controlling the formation of precipitates, which is important in various industries and laboratory practices. For instance, in wastewater treatment, controlling the concentration of ions can prevent the formation of unwanted precipitates that can clog pipes or harm aquatic life.