A solution contains 1.0 \(\times 10^{-5}\) mol of KBr and 0.10 mol of KCl per liter. AgNO \(_{3}\) is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?

Short Answer

Expert verified
Solid AgBr forms first if its required silver ion concentration to reach its solubility product is lower than that of AgCl.

Step by step solution

01

Write down the solubility product constants

Find the solubility product constants from a reference source. The relevant constants are the solubility products for AgBr and AgCl, denoted as \(K_{sp}(AgBr)\) and \(K_{sp}(AgCl)\), respectively.
02

Write the ion equilibrium expressions

Express the solubility products in terms of ionic concentrations for both silver bromide and silver chloride. This can be done using the equilibrium expressions: \(K_{sp}(AgBr) = [Ag^+][Br^-]\) and \(K_{sp}(AgCl) = [Ag^+][Cl^-]\).
03

Calculate ion concentrations when solids begin to form

Use the initial ion concentrations in the solution to find the concentrations when the respective solids begin to form. For AgBr: \([Ag^+]_{AgBr} = \frac{K_{sp}(AgBr)}{[Br^-]}\), and for AgCl: \([Ag^+]_{AgCl} = \frac{K_{sp}(AgCl)}{[Cl^-]}\).
04

Compare the calculated silver ion concentrations

Compare the silver ion concentrations required to start precipitating AgBr and AgCl. The solid with the lower required silver ion concentration forms first.
05

Determine which solid forms first

Based on the calculated silver ion concentrations, determine whether solid AgBr or solid AgCl forms first. The one that requires a lower concentration of silver ions to reach its solubility product will precipitate first.

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Key Concepts

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

Chemical Equilibrium
In the context of solubility, chemical equilibrium refers to a state where the rate of dissolution (the process of a substance dissolving) of a substance is equal to the rate of precipitation (the process of a substance coming out of solution as a solid). This equilibrium is significant in the solubility product constant (Ksp) exercises, such as determining whether AgBr or AgCl will precipitate first.

The Ksp is a reflection of the equilibrium state for a sparingly soluble compound and is specific for any given compound at a particular temperature. The calculation of ionic concentrations at equilibrium is central to predicting the point at which a solid will begin to form in a solution containing multiple ions.

Understanding chemical equilibrium allows us to apply Le Chatelier's Principle. This principle posits that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In our exercise, this means that adding silver ions (Ag+) will shift the equilibrium to form more solid product if it surpasses the equilibrium concentration governed by Ksp.
Ionic Concentration
Ionic concentration is a measure of how many ions of a particular type are present in a solution, typically expressed in moles per liter (Molarity, M). In our example involving KBr, KCl, and AgNO3, ion concentrations are pivotal to solving for which salt will precipitate first. We need to know the initial concentrations of bromide (Br) and chloride (Cl) ions to understand how much Ag+ can be added before reaching the solubility limit, as represented by the Ksp value.

When we talk about ionic concentration in a solubility context, we are often referring to the ions that come from a salt dissociating in water. Considering the equation for the solubility product constant, knowing the ionic concentrations of either reactant or product will allow us to calculate the concentration of the other in equilibrium conditions. The exact molarity values give us the tool to predict and calculate when and which compounds will precipitate out of the solution.
Precipitation Reactions
Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products. This reaction is essential for determining which compound, AgBr or AgCl, precipitates first. Typically, a precipitate will form when the product of the concentrations of the ions exceeds the solubility product constant (Ksp).

For example, when AgNO3 is added to the solution containing KBr and KCl, silver ions combine with bromide and chloride ions. If the product of these ionic concentrations surpasses the respective Ksp value, it indicates that the system has exceeded the maximum concentration of ions that can stay dissolved, and thus a precipitate will form.

To predict the outcome of precipitation reactions, one must compare the relative solubility (Ksp) of the potential precipitates. In our scenario, we compare the ion concentrations at which AgBr and AgCl will begin to precipitate. The one that forms a solid at the lower concentration of silver ions is the one that will precipitate first when Ag+ is added to the solution.

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Most popular questions from this chapter

The Handbook of Chemistry and Physics (http://openstaxcollege.org///16Handbook) gives solubilities of the following compounds in grams per \(100 \mathrm{mL}\) of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) \(\mathrm{BaSeO}_{4}, 0.0118 \mathrm{g} / 100 \mathrm{mL}\) (b) \(\mathrm{Ba}\left(\mathrm{BrO}_{3}\right)_{2} \cdot \mathrm{H}_{2} \mathrm{O}, 0.30 \mathrm{g} / 100 \mathrm{mL}\) (c) \(\mathrm{NH}_{4} \mathrm{MgAsO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}, 0.038 \mathrm{g} / 100 \mathrm{mL}\) (d) \(\mathrm{La}_{2}\left(\mathrm{MoO}_{4}\right)_{3}, 0.00179 \mathrm{g} / 100 \mathrm{mL}\)

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 \(\mathrm{BF}_{3}\).

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{PbBr}_{2}\) (c) AgI (d) \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)

How many grams of Milk of Magnesia, Mg(OH) \(_{2}(s)\) (58.3 g/mol), would be soluble in 200 mL of water. \(K_{\mathrm{sp}}=\) \(7.1 \times 10^{-12} .\) Include the ionic reaction and the expression for \(K_{\mathrm{sp}}\) in your answer. \(\left(K_{\mathrm{w}}=1 \times 10^{-14}=\right.\) \(\left.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\right)\)

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