Question

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}{ }_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Short answer

(a) No, \(\mathrm{Ca}(\mathrm{OH})_{2}\) will not precipitate from the solution, as the calculated ion product \(\mathrm{Q} \le K_{sp}\). (b) Yes, \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) will precipitate when the solutions are mixed, as the calculated ion product \(\mathrm{Q} > K_{sp}\).

Step-by-step solution

Part (a): Calculate the ion product for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Firstly, we need to know the relationship between pH and hydroxide ion concentration: \[\mathrm{pOH} = 14 - \mathrm{pH}\] Calculate the hydroxide ion concentration using the given pH: \[\mathrm{OH^{-}}\,\mathrm{concentration} = 10^{-\mathrm{pOH}}\]

Part (a): Compare the ion product with the solubility product constant

The solubility product constant for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(K_{sp} = 4.68 \times 10^{-6}\). Calculate the ion product, \(\mathrm{Q}\): \[\mathrm{Q} = [\mathrm{Ca^{2+}}][\mathrm{OH^{-}}]^2\] Use the initial concentration of \(\mathrm{CaCl}_{2}\) for the \(\mathrm{Ca^{2+}}\) concentration (as 1 mole of \(\mathrm{CaCl}_{2}\) produces 1 mole of \(\mathrm{Ca^{2+}}\)); and use the calculated hydroxide ion concentration for \(\mathrm{OH^{-}}\). No precipitation occurs if \(\mathrm{Q} \le K_{sp}\), while precipitation occurs if \(\mathrm{Q} > K_{sp}\). Compare the ion product \(\mathrm{Q}\) with the solubility product constant \(K_{sp}\).

Part (b): Calculate the concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions after mixing solutions

Use the dilution formula to find the new concentrations of ions after mixing: \(C_1 V_1 = C_2 V_2\) Calculate new concentrations for \(\mathrm{Ag}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions after mixing the solutions.

Part (b): Compare the ion product with the solubility product constant

The solubility product constant for \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) is \(K_{sp} = 1.2 \times 10^{-5}\). Calculate the ion product, \(\mathrm{Q}\): \[\mathrm{Q} = [\mathrm{Ag^{+}}]^2 [\mathrm{SO_{4}^{2-}}]\] Use the calculated concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions after mixing. No precipitation occurs if \(\mathrm{Q} \le K_{sp}\), while precipitation occurs if \(\mathrm{Q} > K_{sp}\). Compare the ion product \(\mathrm{Q}\) with the solubility product constant \(K_{sp}\).

Key concepts

Solubility Product Constant (Ksp)

When it comes to understanding how a solute dissolves in a solvent, the solubility product constant, often abbreviated as Ksp, is a pivotal concept. It's a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. It gives us a quantitative measure of how far a reaction can proceed before reaching equilibrium and can be used to predict whether a precipitate will form in a given solution.

For a general salt represented as AB, which dissociates into A+ and B- ions, the solubility product is expressed as:Ksp = [A+] [B-].It includes the concentrations of the ions at equilibrium. If you attempt to dissolve more salt than the Ksp value allows, you will get a precipitate as the excess ions form a solid to maintain the equilibrium.

To use Ksp in calculations, you first need to know the formula of the salt and its dissociation products. Then, by substituting the concentrations of these ions into the Ksp expression, you can determine if a solution is saturated (at equilibrium), undersaturated (no precipitate), or supersaturated (precipitate forms).The Ksp value is unique to each compound and typically depends on the temperature; it's a constant at a given temperature. Knowing the solubility product constant can help predict the outcome of mixing solutions containing ions that could form a precipitate. This ability to forecast the solubility is essential in many areas of chemistry, including environmental science, pharmaceuticals, and materials science.

pH and pOH Relationship

The pH and pOH of a solution are a measure of its acidity and basicity, respectively. They provide important information about the concentration of hydrogen ions (H+) and hydroxide ions (OH-) in a solution. The pH and pOH are related by the simple relationship:pH + pOH = 14.This equation holds true for aqueous solutions at 25°C (298 K) and reflects the self-ionization of water where the product of [H+] and [OH-] is a constant value.When you know the pH of a solution, you can easily calculate the pOH by subtracting the pH from 14. This is extremely useful in solving precipitation problems because it allows you to find out the concentration of OH- ions (which you'll need for Ksp calculations) simply by using the relationship: [OH-] = 10^-pOH.The pH scale typically ranges from 0 to 14, with 7 being neutral pH. Values below 7 indicate an acidic solution, while values above 7 suggest a basic one. Having a firm grasp of the pH and pOH relationship is crucial for calculating the ion concentration in solutions, which in turn is necessary for understanding and predicting the reactions in those solutions, such as precipitation events.

Ion Concentration Calculation

Determining ion concentration is a key skill in chemistry that enables you to predict and understand various chemical reactions. When mixing solutions in a precipitation reaction, calculating the concentration of ions after mixing can tell you whether a precipitate will form or not.

A typical approach to find the final ion concentration after mixing involves two steps: the first being the use of the dilution formula:C1V1 = C2V2,where C1 and V1 are the concentration and volume of the first solution, and C2 and V2 are those of the second solution after mixing.

Once you know the new concentrations of ions in the solution, you combine them according to the stoichiometry of the possible precipitate to find the ion product (Q). This is compared to the known solubility product constant (Ksp) to determine if a precipitation reaction will occur. If Q > Ksp, a precipitate forms; if Q <= Ksp, no precipitate forms.

Ion concentration calculations are used not just in academic problems but also in real-world applications such as environmental engineering, where they are used to assess the potential for the formation of scale in pipes, or in medicine, for monitoring the ion concentration in body fluids.