Question

Water from a well is found to contain \(3.0 \mathrm{mg}\) of calcium ion per liter. If \(0.50 \mathrm{mg}\) of sodium sulfate is added to one liter of the well water without changing its volume, will a precipitate form? What should \(\left[\mathrm{SO}_{4}{ }^{2-}\right]\) be to just start precipitation?

Short answer

Answer: Precipitation would begin at a sulfate ion concentration of 3.21 x 10^-4 M.

Step-by-step solution

Determine the \(K_{sp}\) of \(\mathrm{CaSO}_4\)

The solubility product constant (\(K_{sp}\)) for calcium sulfate (\(\mathrm{CaSO}_4\)) is available in textbooks or can be found online. In this case, \(K_{sp}(\mathrm{CaSO}_4) = 2.4 \times 10^{-5}\).

Calculate the initial concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_4^{2-}\)

First, convert calcium ion concentration from mg/L to moles per liter (molarity): \(\left[\mathrm{Ca}^{2+}\right]=(\frac{3.0}{1000})\times\frac{1}{\text{molar mass of Ca}^{2+}}\) The molar mass of \(\mathrm{Ca}^{2+}\) is \(40.08 \mathrm{g/mol}\). So, the initial concentration of \(\mathrm{Ca}^{2+}\) is: \(\left[\mathrm{Ca}^{2+}\right]=(\frac{3.0}{1000})\times\frac{1}{40.08}=7.49\times10^{-5} \mathrm{M}\) Now let's calculate the initial concentration of \(\mathrm{SO}_4^{2-}\) after the addition of 0.50 mg of sodium sulfate. \(\left[\mathrm{SO}_4^{2-}\right]=(\frac{0.50}{1000})\times\frac{1}{\text{molar mass of \)\mathrm{Na_2SO_4}\(}}\times 2\) The molar mass of \(\mathrm{Na_2SO_4}\) is \(142.04 \mathrm{g/mol}\). So, the initial concentration of \(\mathrm{SO}_4^{2-}\) is: \(\left[\mathrm{SO}_4^{2-}\right]=(\frac{0.50}{1000})\times\frac{1}{142.04}\times 2=7.04\times10^{-6} \mathrm{M}\)

Check if the precipitation occurs

In order to determine if a precipitate will form, we must compare the product of the initial concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_4^{2-}\) with the solubility product, \(K_{sp}\). We will compare \((\left[\mathrm{Ca}^{2+}\right]\times\left[\mathrm{SO}_4^{2-}\right])\) with \(K_{sp}\)(Calcium sulfate) i.e \(2.4 \times 10^{-5}\). \((\left[\mathrm{Ca}^{2+}\right]\times\left[\mathrm{SO}_4^{2-}\right])= (7.49\times10^{-5})(7.04\times10^{-6})=5.27\times10^{-10}\) Since \((\left[\mathrm{Ca}^{2+}\right]\times\left[\mathrm{SO}_4^{2-}\right]) < K_{sp}(\mathrm{CaSO}_4)\), \(5.27\times10^{-10} < 2.4 \times 10^{-5}\), no precipitation will occur initially upon addition of sodium sulfate.

Calculate the required \(\left[\mathrm{SO}_4^{2-}\right]\) to just start precipitation

To calculate the concentration of sulfate ions required to just start precipitation, we should set \((\left[\mathrm{Ca}^{2+}\right]\times\left[\mathrm{SO}_4^{2-}\right])\) equal to \(K_{sp}(\mathrm{CaSO}_4)\) So, \((\left[\mathrm{Ca}^{2+}\right]\times\left[\mathrm{SO}_4^{2-}\right]) = 2.4 \times 10^{-5}\) We already have the initial concentration of \(\mathrm{Ca}^{2+}\), which is \(7.49\times10^{-5} \mathrm{M}\). We can then solve for the required concentration of \(\mathrm{SO}_4^{2-}\): \(\Rightarrow \left[\mathrm{SO}_4^{2-}\right] = \frac{K_{sp}}{\left[\mathrm{Ca}^{2+}\right]} = \frac{2.4 \times 10^{-5}}{7.49\times10^{-5}}\) = \(3.21\times10^{-4} \mathrm{M}\) So, to just start precipitation, the concentration of sulfate ions should be \(3.21\times10^{-4} \mathrm{M}\).

Key concepts

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in solution combine to form an insoluble compound, or precipitate. This is fundamentally a type of double displacement reaction where the anions and cations switch partners.For a precipitation reaction to take place, the product of the reaction must be a solid compound that separates from the solution. These reactions are significant in various fields including chemistry, environmental science, and industry. For example, they are important in water treatment to remove undesirable ions from water.The conditions for a precipitation reaction are governed by the solubility rules, which predict whether a compound will be soluble or insoluble in water. However, the quantitative aspect of precipitate formation is described by the solubility product constant (K_{sp}). This constant relates to the maximum amount of the substance that can dissolve in a solution before it starts precipitating.

Chemical Equilibrium

Chemical equilibrium represents a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of the reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring at an equal rate.In the context of solubility, equilibrium is achieved when a saturated solution is in contact with undissolved solute. The K_{sp}, or solubility product constant, is specific to the saturated solution of a sparingly soluble compound and represents the product of the molar concentrations at equilibrium.For example, calcium sulfate in water reaches solubility equilibrium when the product of the calcium ion ({Ca}^{2+}) concentration and the sulfate ion ({SO}_4^{2-}) concentration equals the K_{sp} of calcium sulfate. Any changes in the ion concentrations can shift the equilibrium, potentially leading to the formation of a precipitate if the product exceeds the K_{sp}.

Molar Concentration

Molar concentration, also called molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution. The unit for molar concentration is moles per liter (M) or mol/L. Molar concentration is a fundamental concept in chemistry, especially in the discussion of reactions in solution.Calculating molar concentrations is crucial for predicting whether a precipitate will form in a reaction. By comparing the product of the molar concentrations of the ions involved in the reaction to the K_{sp} value, we can determine if the conditions are met for a precipitation reaction. It is essential to convert all given concentrations into molar concentrations to accurately apply the solubility product constant in quantitative analysis.In the context of the exercise, understanding the initial molar concentrations of each ion involved ({Ca}^{2+} and {SO}_4^{2-}) allows us to predict the formation of a precipitate, by comparing the product of these concentrations to the K_{sp} for calcium sulfate.