Question

The \(K_{\text {sp }}\) for silver sulfate \(\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\) is \(1.2 \times 10^{-5} .\) Calculate the solubility of silver sulfate in each of the following. a. water b. \(0.10 \mathrm{MAgNO}_{3}\) c. \(0.20 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\)

Short answer

The solubility of silver sulfate (Ag₂SO₄) in the given solutions are as follows: a. In water: \( \approx 1.12 \times 10^{-2} M\) b. In 0.1 M AgNO₃: \( \approx 1.2 \times 10^{-3} M\) c. In 0.20 M K₂SO₄: \( \approx 3.87 \times 10^{-3} M\)

Step-by-step solution

Write the chemical equation and the expression for Kᵪ

First, let's write the chemical equation for the dissolution of silver sulfate: \(Ag_2SO_4(s) \rightleftharpoons 2Ag^+(aq) + SO_4^{2-}(aq)\) Now let's write the expression for Kᵪ: \(K_{sp} =[Ag^+]^2[SO_4^{2-}]\) Given \(K_{sp} = 1.2 \times 10^{-5}\)

Calculate the solubility of Ag₂SO₄ in water

We can set up an ICE (Initial, Change, Equilibrium) table for the dissolution of Ag₂SO₄: | | 2Ag⁺ | SO₄²⁻ | | ------ | ---- | ----- | | Initial | 0 | 0 | | Change | +2s | +s | | Equil | 2s | s | Plugging these concentration values into the Kᵪ expression: \(1.2 \times 10^{-5} =(2s)^2(s)\) Now solve for s: \(s^3 = \frac{1.2\times 10^{-5}}{4}\) \(s = \sqrt[3]{\frac{1.2\times 10^{-5}}{4}}\) \(s \approx 1.12 \times 10^{-2}M\) The solubility of silver sulfate in water is approximately 1.12 x 10² M.

Calculate the solubility of Ag₂SO₄ in 0.1 MAgNO₃

In this case, we will have an initial concentration of Ag⁺ from the AgNO₃: | | 2Ag⁺ | SO₄²⁻ | | ------ | ---------- | ----- | | Initial | 0.1 | 0 | | Change | +2s | +s | | Equil | 0.1+2s | s | Plugging these concentration values into the Kᵪ expression: \(1.2 \times 10^{-5} =(0.1 + 2s)^2(s)\) Since 2s is very small, we can simplify it to: \(1.2 \times 10^{-5} =(0.1)^2(s)\) Now solve for s: \(s = \frac{1.2\times 10^{-5}}{0.1^2}\) \(s \approx 1.2 \times 10^{-3} M\) The solubility of silver sulfate in 0.1 M AgNO₃ is approximately 1.2 x 10⁻³ M.

Calculate the solubility of Ag₂SO₄ in 0.20 MK₂SO₄

In this case, we will have an initial concentration of SO₄²⁻ from the K₂SO₄: | | 2Ag⁺ | SO₄²⁻ | | ------ | ---- | --------------- | | Initial | 0 | 0.2 | | Change | +2s | +s | | Equil | 2s | 0.2 + s | Plugging these concentration values into the Kᵪ expression: \(1.2 \times 10^{-5} =(2s)^2(0.2+s)\) Since s is very small, we can simplify it to: \(1.2 \times 10^{-5} =(2s)^2(0.2)\) Now solve for s: \(s = \sqrt{\frac{1.2\times 10^{-5}}{0.2\times4}}\) \(s \approx 3.87 \times 10^{-3} M\) The solubility of silver sulfate in 0.2 M K₂SO₄ is approximately 3.87 x 10⁻³ M.

Key concepts

Understanding Silver Sulfate Solubility

Silver sulfate (Ag_2SO_4) is a chemical compound that possesses a certain solubility in water, which refers to its ability to dissolve to form a homogenous solution. The solubility of silver sulfate is influenced by the presence of other compounds in the solution, which can provide ions that interact with the silver or sulfate ions. This concept is particularly important when predicting how much of \(Ag_2SO_4\) will dissolve in different environments, such as pure water or solutions containing other silver or sulfate salts.

Exercise Improvement Advice:

When calculating solubility using the solubility product constant (K_{sp}), remember to consider the stoichiometry of the dissociation equation. Each mole of silver sulfate produces two moles of silver ions (Ag^+) and one mole of sulfate ions (SO_4^{2-}), affecting the calculations of their concentrations at equilibrium.

Ksp Calculations Simplified

The solubility product constant, or \(K_{sp}\), quantifies the solubility of a sparingly soluble compound. It is the product of the concentrations of the ions that are produced when the compound dissolves, each raised to the power of its coefficient in the balanced equation. In the given problem, the \(K_{sp}\) for silver sulfate is \(1.2 \times 10^{-5}\). The calculations involve setting up an equation based on the solubility expression and solving for the solubility, denoted as \(s\).

The precision of \(K_{sp}\) calculations heavily relies on acknowledging the ionic dissociation and the correct setup of the equilibrium expression. It's crucial for students to practice this process to gain confidence in their calculations.

Important Note:

When the ionic concentration of one component is known, it can significantly simplify the calculations by assuming that changes in its concentration due to solubility are negligible.

Chemical Equilibrium and Its Role in Solubility

Chemical equilibrium occurs in reversible reactions when the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products. The concept of equilibrium is critical when dealing with the solubility of slightly soluble compounds like silver sulfate. While the solid phase of such compounds does not appear in the equilibrium expression, the product of the concentrations of the dissolved ions at equilibrium is essential in determining the point at which no more solid will dissolve in the solution at a given temperature.

For learners, exploring how chemical equilibrium relates to solubility can provide deeper insights into how changes in conditions, like adding more of a reactant or product, can shift the equilibrium point, thus changing the solubility.

ICE Table Method Explained

The ICE table method stands for Initial, Change, Equilibrium—a tool used to visualize and calculate the concentrations of species in a chemical equilibrium. This method functions as a systematic approach to deal with equilibrium calculations and is particularly useful in solubility exercises.

The 'Initial' row represents the initial concentrations of reactants and products, the 'Change' row shows the changes in concentrations as the system reaches equilibrium, and the 'Equilibrium' row displays the concentrations at equilibrium. Solving the \(K_{sp}\) often involves assuming that the initial concentration of a solid solute is zero, as it is not in solution until it starts to dissolve.

When a common ion is present in the solution, as seen in parts (b) and (c) of the exercise, the ICE table method helps to properly account for the initial ion concentration that impacts the solubility of the solid.