A solution contains \(1.0 \times 10^{-2}\) mol of \(\mathrm{KI}\) and 0.10 mol of \(\mathrm{KCl}\) per liter. AgNO \(_{3}\) is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Short Answer

Expert verified
Solid AgCl forms first because its ion product will reach its solubility product before the ion product of AgI does, due to higher initial concentration of Cl- ions.

Step by step solution

01

- Calculate the solubility product (Ksp)

We need to find the solubility products for AgI and AgCl. The solubility product is given by the product of the molar concentrations of the ions, raised to the power of their stoichiometric coefficients in the equilibrium expression. For AgI, this is given by Ksp(AgI) = [Ag+][I-]. For AgCl, it is Ksp(AgCl) = [Ag+][Cl-].
02

- Determine the molar concentrations

The molar concentrations of I- and Cl- ions in the solution are already given. For I-, it is 1.0 x 10^-2 M. For Cl-, it is 0.10 M.
03

- Use the Ksp values

Look up the Ksp values in a chemistry resource for AgI and AgCl. Common values are around 8.3 x 10^-17 mol2/L2 for AgI and 1.8 x 10^-10 mol2/L2 for AgCl.
04

- Calculate ion product (Q) for each compound

The ion product Q for each silver compound is the product of the molar concentrations of the respective ions. For AgI, Q(AgI) = [Ag+][I-]. For AgCl, Q(AgCl) = [Ag+][Cl-]. Initially, [Ag+] is zero and will increase as AgNO3 is added.
05

- Predict which solid forms first

The solid that will form first is the one whose Q (ion product) equals or exceeds its Ksp first. Since the molar concentration of Cl- is much greater than that of I-, the Q for AgCl will reach the Ksp for AgCl before the Q for AgI reaches the Ksp for AgI. Therefore, AgCl will precipitate before AgI.

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Key Concepts

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

Ksp Calculation
Understanding how to calculate the solubility product constant ({ Ksp}) is crucial in predicting precipitation reactions. It's a straightforward but significant concept in chemistry. The { Ksp} is the product of the concentrations of the ions in a saturated solution, each raised to the power equal to the coefficient of that ion in the dissociation equation.

For instance, when considering silver iodide (AgI), it dissociates into Ag+ and I- ions. The { Ksp} expression for AgI is written as { Ksp(AgI) = [Ag+][I-]}. Here, the square brackets represent the molar concentration of the ions. If the molar concentration of I- is given, and the initial concentration of Ag+ is zero (which changes as a precipitant is added), the { Ksp} can help us determine at which point a solid will start to form.

For a proper { Ksp} calculation, students must diligently follow the steps such as identifying the ionic constituents, determining their molar concentrations, and understanding the stoichiometry of the compound. The ability to calculate { Ksp} enables students to predict the formation of a precipitate under varying solution conditions.
Precipitation Reaction
A precipitation reaction occurs when two soluble salts react in solution to form one or more insoluble products, known as precipitates. This is an essential concept in chemistry, particularly in understanding how ionic compounds interact in solution. When a precipitate forms, it will do so because the product of the concentrations of the ions exceeds the { Ksp} of the compound.

To anticipate which compound precipitates first, as illustrated in the exercise with AgI and AgCl, you compare the ion product (Q) for each compound with its respective { Ksp}. For AgCl, with a higher concentration of Cl- ions, the ion product will increase faster than for AgI when AgNO3 is added. As AgNO3 dissociates into Ag+ and NO3-, the Ag+ ions combine with Cl- until the product of their concentrations reaches or exceeds the { Ksp} of AgCl, leading to the formation of solid AgCl. Understanding the dynamics of precipitation reactions is essential for predicting the outcomes of mixing different ionic solutions.
Chemical Equilibrium
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction. This delicate balance means that the concentrations of reactants and products remain constant over time, as long as the system is closed and conditions are stable.

In the context of solubility, equilibrium exists between a solid compound and its ions in a saturated solution. The { Ksp} is an expression of this equilibrium for sparingly soluble salts. As more precipitant is added, as with AgNO3 in our example, the equilibrium shifts to accommodate the additional Ag+ ions. Should the concentration of the product ions exceed the { Ksp}, the equilibrium will be disturbed, resulting in the formation of a solid precipitate.

Thus, understanding chemical equilibrium is not just about reactions at a standstill but also the factors that can shift the equilibrium in favor of the reactants or products. With this knowledge, students can manipulate reactions to either prevent or encourage the formation of precipitates.

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Most popular questions from this chapter

We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, \(\mathrm{H}^{+}\) ) to another compound is called a Bronsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Bronsted-Lowry definition and the Lewis definition are microscopic definitions.

Assuming that no equilibrium other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) \(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) (b) \(\mathrm{PbI}_{2}\) (c) \(\mathrm{Ag}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), a salt containing the \(\mathrm{Fe}(\mathrm{CN})_{4}^{-}\) ion (d) \(\mathrm{Hg}_{2} \mathrm{I}_{2}\)

The carbonate ion concentration is gradually increased in a solution containing divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will form first? Which of the following will form last? Explain. \(\begin{array}{ll}\text { (a) } \mathrm{MgCO}_{3} & K_{\mathrm{sp}}=3.5 \times 10^{-8}\end{array}\) (b) \(\mathrm{CaCO}_{3} \quad K_{\mathrm{sp}}=4.2 \times 10^{-7}\) (c) \(\operatorname{Sr} \mathrm{CO}_{3} \quad K_{\mathrm{sp}}=3.9 \times 10^{-9}\) \(\begin{array}{ll}\text { (d) } \mathrm{BaCO}_{3} & K_{\mathrm{sp}}=4.4 \times 10^{-5}\end{array}\) (e) \(\mathrm{MnCO}_{3} \quad K_{\mathrm{sp}}=5.1 \times 10^{-9}\)

In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point \(=19.5^{\circ} \mathrm{C}\) ) is a strong acid. In liquid HF, HNO _ acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept \(F^{-}\) ion (for example, \(B F_{3}\) or \(S\) bF \(_{5}\) ). Write balanced chemical equations for the reaction of pure \(\mathrm{HNO}_{3}\) with pure \(\mathrm{HF}\) and of pure \(\mathrm{HF}\) with \(\mathrm{BF}_{3}\).

What is the effect on the amount of CaHPO \(_{4}\) that dissolves and the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HPO}_{4}^{-}\) -when each of the following are added to a mixture of solid \(\mathrm{CaHPO}_{4}\) and water at equilibrium? (a) \(\mathrm{CaCl}_{2}\) (b) HCl (c) \(\mathrm{KClO}_{4}\) (d) \(\mathrm{NaOH}\) (e) \(\mathrm{CaHPO}_{4}\)

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