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Chapter 1: Problem 31

Silver sulfate, \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) , has a solubility product constant of \(1.0 \times 10^{-5} .\) The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed. (Diagram Can't Copy) Which ion concentrations below would have led the precipitate to form? (A) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (B) \(\left[\mathrm{Ag}^{+}\right]=0.10 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (C) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.10 M\) (D) This is impossible to determine without knowing the total volume of the solution.

Short Answer

Expert verified
The ion concentrations that would have led the silver sulfate precipitate to form are \([\mathrm{Ag}^{+}]=0.10 M\) and \([\mathrm{SO}_{4}^{2-}]=0.01 M\).

Step by step solution

01

Understanding the Solubility Product Expression

To determine the ion concentrations that lead to precipitation, we first write the solubility product expression for silver sulfate, which is \(K_{sp} = [\mathrm{Ag}^{+}]^{2} [\mathrm{SO}_{4}^{2-}]\). This expression is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
02

Substituting the Given Data

Let's substitute the given values into the \(K_{sp}\) expression and see if they would lead to precipitation by comparing with the given \(K_{sp}\) value of \(1.0 \times 10^{-5}\).\n\n(A) For \([\mathrm{Ag}^{+}]=0.01 M\) and \([\mathrm{SO}_{4}^{2-}]=0.01 M\), \(K' = (0.01)^{2} \times 0.01 = 1 \times 10^{-6}\) which is less than \(K_{sp}\), so no precipitation will occur.\n\n(B) For \([\mathrm{Ag}^{+}]=0.10 M\) and \([\mathrm{SO}_{4}^{2-}]=0.01 M\), \(K' = (0.10)^{2} \times 0.01 = 1 \times 10^{-4}\) which is higher than \(K_{sp}\), so precipitation will occur. \n\n(C) For \([\mathrm{Ag}^{+}]=0.01 M\) and \([\mathrm{SO}_{4}^{2-}]=0.10 M\), \(K' = (0.01)^{2} \times 0.10 = 1 \times 10^{-5}\), equals to \(K_{sp}\), so no precipitation, or at equilibrium.
03

Interpreting the Results

By comparing \(K'\) (calculated solubility product) with \(K_{sp}\), we can infer if a precipitate will form or not. If \(K'> K_{sp}\), then the solution is supersaturated and precipitation would occur; if \(K'\leq K_{sp}\), then the solution is unsaturated or at equilibrium and no precipitation would occur. The total volume of the solution is not necessary to determine unless we were asked to calculate the amount of precipitate formed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reaction
A precipitation reaction is a type of chemical reaction where ions in a solution combine to form an insoluble solid, known as the precipitate. This reaction occurs when the product of the ion concentrations in a solution exceeds the solubility product constant, known as the solubility limit. When this limit is crossed, excess ions form an insoluble compound, thus precipitating out of the solution.

In the context of silver sulfate, when the concentrations of silver ions (\([\mathrm{Ag}^+]\)) and sulfate ions (\([\mathrm{SO}_4^{2-}]\)) in a solution multiply to exceed the solubility product constant of silver sulfate, \(K_{sp} = 1.0 \times 10^{-5}\), a solid silver sulfate precipitate will form. Understanding precipitation reactions helps predict whether a mixture of ions will result in a solid forming, important in fields like chemistry and environmental science.
Silver Sulfate
Silver sulfate, \(\mathrm{Ag}_2\mathrm{SO}_4\), is a white crystalline solid that is only slightly soluble in water. Its solubility is described by a constant, known as the solubility product constant (\(K_{sp}\)), which for silver sulfate is \(1.0 \times 10^{-5}\). This constant provides essential information on the solubility of the compound in water.

The lower the \(K_{sp}\) value, the less soluble the compound is under normal conditions. With silver sulfate's low value, it readily forms a precipitate when the conditions permit. Understanding the behavior of silver sulfate's ions in a solution helps predict and control precipitation reactions, which is crucial when working with solutions containing this compound.
Ion Concentrations
Ion concentrations in a solution are critical for determining whether a reaction will form a precipitate. The concentration is usually expressed in molarity (M), symbolized as \([\mathrm{M}]\). When dealing with precipitation reactions, the product of the concentrations of the ions involved is compared to the solubility product constant, \(K_{sp}\).

For silver sulfate (\(\mathrm{Ag}_2\mathrm{SO}_4\)), the reaction can form a precipitate if \([\mathrm{Ag}^+]^2[\mathrm{SO}_4^{2-}] > K_{sp}\). By calculating this product, one can predict the possibility of precipitation:
  • If the calculated value exceeds the \(K_{sp}\), the solution is supersaturated, and a precipitate forms.
  • If it is equal to or less than the \(K_{sp}\), the solution is at equilibrium, or not saturated enough to form a precipitate.
Understanding ion concentration and its impact is vital for chemists and scientists addressing reactions in aqueous solutions.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction proceeds at the same rate in both directions; meaning, the concentrations of products and reactants remain unchanged over time. In a solution involving precipitation, equilibrium is when the rate of the ions combining to form the solid equals the rate of the solid dissolving back into ions.

For silver sulfate, equilibrium is particularly interesting when calculating \([\mathrm{Ag}^+]^2[\mathrm{SO}_4^{2-}] = K_{sp}\), exactly matches the solubility product constant. At this point, no new precipitate will form, because the system is balanced. Recognizing chemical equilibrium is fundamental for understanding the dynamics of solutions in chemistry, including the conditions under which reactions will stabilize without further changes in state.

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Most popular questions from this chapter

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

\(2 \mathrm{HI}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{I}_{2}(g)+\) energy A gaseous reaction occurs and comes to equilibrium, as shown above. Which of the following changes to the system will serve to increase the number of moles of \(\mathrm{I}_{2}\) present at equilibrium? (A) Increasing the volume at constant temperature (B) Decreasing the volume at constant temperature (C) Increasing the temperature at constant volume (D) Decreasing the temperature at constant volume

$$2 \mathrm{NOCl} \rightarrow 2 \mathrm{NO}+\mathrm{Cl}_{2}$$ The reaction above takes place with all of the reactants and products in the gaseous phase. Which of the following is true of the relative rates of disappearance of the reactants and appearance of the products? (A) NO appears at twice the rate that NOCl disappears. (B) NO appears at the same rate that NOCl disappears. (C) NO appears at half the rate that NOCl disappears. (D) \(\mathrm{Cl}_{2}\) appears at the same rate that NOCl disappears.

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)
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