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A solution contains three anions with the following concentrations: \(0.20 \mathrm{MCrO}_{4}^{2-}, 0.10 \mathrm{MCO}_{3}^{2-}\), and \(0.010 \mathrm{M} \mathrm{Cl}^{-}\). If a dilute \(\mathrm{AgNO}_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{3 p}=1.2 \times 10^{-12}\right)\). \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{\mathrm{p}}=8.1 \times 10^{-12}\right)\), or \(\mathrm{AgCl}\left(K_{\text {sp }}=1.8 \times 10^{-10}\right)\) ?

Short Answer

Expert verified
The first compound to precipitate is Ag鈧侰rO鈧 when a dilute AgNO鈧 solution is added to the given solution, as it has the smallest silver ion concentration (\(4.9 \times 10^{-6} \mathrm{M}\)) at which Q = Ksp.

Step by step solution

01

Write the balanced equations for the formation of silver salts

Write the balanced equations for the precipitation of each of the silver salts: 1. Ag鈧侰rO鈧: 2Ag鈦 + CrO鈧劼测伝 鈫 Ag鈧侰rO鈧 2. Ag鈧侰O鈧: 2Ag鈦 + CO鈧兟测伝 鈫 Ag鈧侰O鈧 3. AgCl: Ag鈦 + Cl鈦 鈫 AgCl
02

Calculate the reaction quotient (Q) for each reaction

For each reaction, use the concentrations of the ions given in the problem to calculate Q: 1. \(Q_{Ag_2CrO_4} = [Ag^+]^2[CrO_4^{2-}]\) 2. \(Q_{Ag_2CO_3} = [Ag^+]^2[CO_3^{2-}]\) 3. \(Q_{AgCl} = [Ag^+][Cl^-]\) Assume that the silver ion concentration ([Ag鈦篯) is the same for all reactions, as it is contributed by the common AgNO鈧. So, we can denote [Ag鈦篯 as x. Now, substitute the given anion concentrations and represent Q for each reaction. 1. \(Q_{Ag_2CrO_4} = x^2(0.20)\) 2. \(Q_{Ag_2CO_3} = x^2(0.10)\) 3. \(Q_{AgCl} = x(0.010)\)
03

Compare the Ksp values with Q for each reaction

Now we will compare the Ksp values with Q for each reaction. Find x such that Q = Ksp for each reaction. Whichever reaction has the smallest x will be the first to precipitate. 1. \(Q_{Ag_2CrO_4} = x^2(0.20) = 1.2 \times 10^{-12}\) 鈫 \(x = \sqrt{\frac{1.2 \times 10^{-12}}{0.20}}\) 2. \(Q_{Ag_2CO_3} = x^2(0.10) = 8.1 \times 10^{-12}\) 鈫 \(x = \sqrt{\frac{8.1 \times 10^{-12}}{0.10}}\) 3. \(Q_{AgCl} = x(0.010) = 1.8 \times 10^{-10}\) 鈫 \(x = \frac{1.8 \times 10^{-10}}{0.010}\) Calculate the value of x for each reaction: 1. \(x_{Ag_2CrO_4} = \sqrt{\frac{1.2 \times 10^{-12}}{0.20}} = 4.9 \times 10^{-6} \mathrm{M}\) 2. \(x_{Ag_2CO_3} = \sqrt{\frac{8.1 \times 10^{-12}}{0.10}} = 9.0 \times 10^{-6} \mathrm{M}\) 3. \(x_{AgCl} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \mathrm{M}\) Since \(x_{Ag_2CrO_4}\) is the smallest, Ag鈧侰rO鈧 will be the first compound to precipitate.

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Key Concepts

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

Solubility Product Constant
Understanding the solubility product constant, denoted as Ksp, is critical when exploring precipitation reactions. This constant represents the equilibrium between a solid and its respective ions in solution. At this dynamic equilibrium, the rate at which the solid dissolves is equal to the rate at which it precipitates out of solution.

The Ksp is unique for every slightly soluble ionic compound and depends on the temperature. For a general compound, represented as \(A_mB_n\), which dissolves into \(m\) A+ ions and \(n\) B- ions, the Ksp expression is given by:
\[K_{\text{sp}} = [A^+]^m[B^-]^n\]
It's imperative to notice that only the ions' concentrations are part of the expression, and the solids are not included. The lower the Ksp value, the less soluble the compound is. In the context of the exercise, Ksp values help us predict which compound will precipitate first by comparing these values with the reaction quotient, Q.
Reaction Quotient
The reaction quotient (Q) is a mathematical expression that compares the concentrations of the reactants and products at any point in time during a reaction. It has the same form as the equilibrium constant expression but differs because it doesn't necessarily reflect the system at equilibrium.

For the precipitation reactions involved in our discussion, Q is assessed by calculating the product of the relevant ion concentrations raised to the power of their stoichiometric coefficients. For instance:
\[Q = [A^+]^m[B^-]^n\]

Comparing Q to Ksp

When Q < Ksp, the reaction tends to form more solid, indicating undersaturation. If Q > Ksp, the reaction moves towards forming more ions, meaning the solution is oversaturated and precipitation is likely to occur. At Q = Ksp, the system is at equilibrium, showing no net change in the amounts of solids or ions.
Ionic Equilibria
Ionic equilibria refer to the state where the rates of dissolution and precipitation of an ionic solid in a solution are equal, creating a dynamic but stable system. These reactions are reversible and can be represented as:
\[A_mB_n(s) \rightleftharpoons mA^+(aq) + nB^-(aq)\]
Here, \(A_mB_n\) is a slightly soluble ionic compound. The solubility product (Ksp) is a key concept for understanding ionic equilibria as it quantifies the maximum product of the ions that can exist in a solution at equilibrium. The equilibrium can shift in response to changes in concentration, temperature, or the presence of common ions, a principle known as Le Chatelier's Principle.
Silver Salts Precipitation
Silver salts such as Ag鈧侰rO鈧, Ag鈧侰O鈧, and AgCl are known for their varying solubilities in water. In the context of our problem, we were tasked with identifying which silver salt would precipitate first upon the gradual addition of AgNO鈧 to a solution containing different anions.

Since precipitation occurs when the ion product exceeds the solubility product constant, we evaluated the Q for each potential silver salt given the starting ion concentrations in the solution. This approach allows us to predict the order of precipitation based on which ion product first reaches or surpasses their respective Ksp value.

Consequently, due to its relatively low Ksp and derived concentrations, Ag鈧侰rO鈧 would be the first to precipitate, indicating a highly insoluble nature among the given silver salts and suggesting a strategy for separation and identification of ions in a mixture. This is a classic case demonstrating the principles of solubility and precipitation in action, important concepts in analytical and inorganic chemistry.

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Most popular questions from this chapter

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (e) \(\mathrm{CuBr}\).

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH})\), (b) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right)\), (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\right)\).

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a \(50.00-\mathrm{mL}\) sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M} \mathrm{HQ}\) to reach the end point. Calculate \(K_{\mathrm{wp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\). Compare your result with that in Appendix D. \(25^{\circ} \mathrm{C}\). Suggest a reason for any differences you find between your value and the one in Appendix D.

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{\mathrm{s}}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}_{\text {k }}\) whose \(K_{p}=1.0 \times 10^{-40}\), (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Calculate the solubility of \(\mathrm{Mg}_{\mathrm{g}}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH} \mathrm{NCl}_{4}\).

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