Question

The $K_{\mathrm{sp}}$ 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. $0.10M$ AgNO ${}_3$ c. $0.20M{\mathrm{K}}_2{\mathrm{SO}}_4$

Short answer

The solubility of silver sulfate (Ag₂SO₄) in the given solutions are: a. In water, the solubility is $3.0\times {10}^{-2}\mathrm{M}$. b. In 0.10 M AgNO₃, the solubility is approximately $1.2\times {10}^{-4}\mathrm{M}$. c. In 0.20 M K₂SO₄, the solubility is approximately $1.2\times {10}^{-3}\mathrm{M}$.

Step-by-step solution

Write the dissolution equation and equilibrium expression

Begin by writing the balanced equation for the dissolution of silver sulfate in water: $${\mathrm{Ag}}_2{\mathrm{SO}}_4(\mathrm{s})\rightleftharpoons 2{\mathrm{Ag}}^{+}(\mathrm{aq})+{\mathrm{SO}}_4^{2-}(\mathrm{aq})$$ Now, write the equilibrium expression for Ksp using the concentrations of the ions: $${\mathrm{K}}_{sp}=[{\mathrm{Ag}}^{+}{]}^2[{\mathrm{SO}}_4^{2-}]$$

Set up an ICE table and solve for solubility

Assume an S molar amount of silver sulfate dissolve in water. Set up an ICE (Initial, Change, Equilibrium) table to help determine the solubility: Initial: | 0 | 0 Change: | +2S | +S Equilibrium: | 2S | S Plug the equilibrium values into the equilibrium expression: $${\mathrm{K}}_{sp}=(2\mathrm{S}{)}^2(\mathrm{S})$$ Now, solve for S: $1.2\times {10}^{-5}=4{\mathrm{S}}^3$ $\mathrm{S}=3.0\times {10}^{-2}\mathrm{M}$ Solubility of silver sulfate in water is $3.0\times {10}^{-2}\mathrm{M}$. #b. Solubility in 0.10 M AgNO₃#

Recognize common ions and write the equilibrium expression

First, note that AgNO₃ shares a common ion (Ag⁺) with silver sulfate. Rewrite the equilibrium expression: $${\mathrm{K}}_{sp}=[{\mathrm{Ag}}^{+}][{\mathrm{SO}}_4^{2-}]$$

Set up an ICE table and solve for solubility

Again, use the ICE table, but this time include the initial concentration of Ag⁺ from AgNO₃ (0.10 M): Initial: | 0.10 M | 0 Change: | +S | +S Equilibrium: | 0.10 + S | S Plug the equilibrium values into the equilibrium expression: $1.2\times {10}^{-5}=(0.10+\mathrm{S})(\mathrm{S})$ Since S will be much smaller than 0.10 M, we can approximate the equation to: $1.2\times {10}^{-5}\approx (0.10)(\mathrm{S})$ $\mathrm{S}\approx 1.2\times {10}^{-4}\mathrm{M}$ Solubility of silver sulfate in 0.10 M AgNO₃ is approximately $1.2\times {10}^{-4}\mathrm{M}$. #c. Solubility in 0.20 M K₂SO₄#

Recognize common ions and write the equilibrium expression

K₂SO₄ shares a common ion (SO₄²⁻) with silver sulfate. Rewrite the equilibrium expression: $${\mathrm{K}}_{sp}[{\mathrm{Ag}}^{+}][{\mathrm{SO}}_4^{2-}]$$

Set up an ICE table and solve for solubility

Again, use the ICE table, but this time include the initial concentration of SO₄²⁻ from K₂SO₄ (0.20 M): Initial: | 0 | 0.20 M Change: | +2S | +S Equilibrium: | 2S | 0.20 + S Plug the equilibrium values into the equilibrium expression: $1.2\times {10}^{-5}=(2\mathrm{S}{)}^2(0.20+\mathrm{S})$ Since S will be much smaller than 0.20 M, we can approximate the equation to: $1.2\times {10}^{-5}\approx (4{\mathrm{S}}^2)(0.20)$ $\mathrm{S}\approx \sqrt{\frac{1.2\times {10}^{-5}}{0.8}}$ $\mathrm{S}\approx 1.2\times {10}^{-3}\mathrm{M}$ Solubility of silver sulfate in 0.20 M K₂SO₄ is approximately $1.2\times {10}^{-3}\mathrm{M}$.

Key concepts

Ksp (Solubility Product Constant)

The concept of the solubility product constant, or Ksp, is a fundamental aspect in understanding the solubility of ionic compounds in solution. This value gives us a measure of the product of the molar concentrations of the ions in a saturated solution, raised to the power of their respective stoichiometric coefficients. As a quantitative expression, Ksp is given by the equation $$K_{sp}=[Cation{]}^m[Anion{]}^n$$where $[Cation]$ and $[Anion]$ are the concentrations of the ions and $m$ and $n$ are their coefficients in the balanced dissolution equation. This equilibrium constant is only applicable for sparingly soluble ionic compounds where the dissolution process reaches a dynamic equilibrium. The lower the value of Ksp, the less soluble the compound is in solution.

In the exercise provided, the Ksp value is used to calculate the solubility of silver sulfate in different scenarios. Employing the formula $K_{sp}=[A{g}^{+}{]}^2[S{O}_4^{2-}]$ for silver sulfate, we find the solubility of the substance in pure water. This process, while intuitive, considers that the system has reached equilibrium, and the product of the concentrations of ions equals the established Ksp value, providing an insight into the maximum amount of substance that can dissolve in water without forming a precipitate.

Common Ion Effect

The common ion effect plays a critical role in the solubility of electrolytes and is a direct consequence of Le Chatelier's principle. This phenomenon occurs when a compound is dissolved in a solution already containing one of the ions present in the compound. The presence of a common ion suppresses the dissolution process, leading to a lower solubility of the compound in such a solution. Essentially, adding a common ion to the solution shifts the equilibrium towards the left, according to the equilibrium expression, causing less of the solid to dissolve.

For example, when silver sulfate is added to a solution containing either AgNO₃ or K₂SO₄, the solubility of silver sulfate decreases because both Ag⁺ and SO₄²⁻ ions are already present in the solution. In the exercise, we observed this effect while calculating the solubility in the presence of Ag⁺ ions from AgNO₃ and SO₄²⁻ ions from K₂SO₄. By incorporating the starting concentration of the common ion into the ICE table, the resulting calculation accounts for the suppressed solubility, demonstrating the impact of the common ion effect on the dissolution process.

Equilibrium Expression

Equilibrium in a chemical reaction is a state where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of reactants and products remain constant over time. The equilibrium expression, derived from the Law of Mass Action, is a mathematical representation of this state. It relates the concentrations of products and reactants to the equilibrium constant ($K$).

In the context of solubility, the equilibrium expression for the solubility product ($K_{sp}$) is exclusively associated with the ions produced from the dissolution of ionic solids. As seen in the solved exercise, the equilibrium expression for the dissolution of silver sulfate is $K_{sp}=[A{g}^{+}{]}^2[S{O}_4^{2-}]$. It is important to emphasize that only species in the aqueous phase are included in the expression, while solids and liquids are omitted. By manipulating this expression, we can predict whether a precipitate will form and calculate the solubility of a sparingly soluble ionic compound in various environments.

ICE Table

An ICE table is an essential tool for solving equilibrium problems, including those involving solubility products. ICE stands for Initial, Change, and Equilibrium and is a methodical way to organize the amounts or concentrations of reactants and products involved in a chemical reaction at different stages. To use it, one lists the initial concentrations of reactants and products, specifies the changes that occur as the system moves towards equilibrium, and then uses these to calculate the equilibrium concentrations.

By setting up an ICE table, you allow yourself to visualize the shift in concentrations of species as the reaction proceeds. This shift is particularly valuable when working with solubility equilibrium, as it helps determine the solubility of the compound. The steps are articulated in the exercise, demonstrating how to complete the ICE table for the dissolution of silver sulfate and taking into account the presence of a common ion.