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Chapter 17: Problem 75

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

Short Answer

Expert verified
AgCl precipitates first.

Step by step solution

01

Understanding Solubility Product

Each potential precipitate has a solubility product constant, or \(K_{sp}\), which indicates its solubility in water. Lower \(K_{sp}\) values typically mean the compound is less soluble and more likely to precipitate first.
02

Writing the Solubility Product Expressions

For each compound, set up the expression for \(K_{sp}\): 1. \( ext{Ag}_2 ext{CrO}_4\): \(K_{sp} = [ ext{Ag}^+]^2[ ext{CrO}_4^{2-}]\)2. \( ext{Ag}_2 ext{CO}_3\): \(K_{sp} = [ ext{Ag}^+]^2[ ext{CO}_3^{2-}]\)3. \( ext{AgCl}\): \(K_{sp} = [ ext{Ag}^+][ ext{Cl}^-]\)
03

Calculating Required [Ag+] for Precipitation

Determine the concentration of \([ ext{Ag}^+]\) needed to start precipitation for each anion:1. For \( ext{Ag}_2 ext{CrO}_4\): \([Ag^+] = \sqrt{\frac{K_{sp}}{[ ext{CrO}_4^{2-}]}} = \sqrt{\frac{1.2 \times 10^{-12}}{0.20}}\)2. For \( ext{Ag}_2 ext{CO}_3\): \([Ag^+] = \sqrt{\frac{K_{sp}}{[ ext{CO}_3^{2-}]}} = \sqrt{\frac{8.1 \times 10^{-12}}{0.10}}\)3. For \( ext{AgCl}\): \([Ag^+] = \frac{K_{sp}}{[ ext{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.010}\)
04

Comparing [Ag+] Values to Determine First Precipitate

Calculate each required \([Ag^+]\):1. \(\text{Ag}_2\text{CrO}_4\): \([Ag^+] = \sqrt{6 \times 10^{-12}} \approx 2.45 \times 10^{-6}\)2. \(\text{Ag}_2\text{CO}_3\): \([Ag^+] = \sqrt{8.1 \times 10^{-11}} \approx 9.0 \times 10^{-6}\)3. \(\text{AgCl}\): \([Ag^+] = 1.8 \times 10^{-8}\)Since the lowest \([Ag^+]\) required for precipitation is for \( ext{AgCl}\), it will precipitate first.

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Key Concepts

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

Precipitation Reactions
Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid, known as a precipitate. This happens due to the formation of a compound that has a low solubility in water. When a solution of \(\text{AgNO}_3\) is added to a solution containing different anions, the silver ions (\(\text{Ag}^+\)) react with available anions to form potential precipitates such as \(\text{Ag}_2\text{CrO}_4\), \(\text{Ag}_2\text{CO}_3\), or \(\text{AgCl}\). Which compound forms first depends on their respective solubility products.
These reactions are often predictable. By measuring the concentration of ions and comparing the product of the ion concentrations to known \(K_{sp}\) values, it's possible to predict whether a precipitate will form. Lower \(K_{sp}\) values suggest compounds are less soluble and more likely to precipitate.
Chemical Equilibrium
Chemical equilibrium in a reaction is reached when the rates of the forward and backward reactions are equal. In the context of precipitation and dissolution, this is when the rate at which the compound dissolves is equal to the rate at which it precipitates out of solution.

For the compounds involved in our precipitation example, the compounds are described by their \(K_{sp}\), which is a type of equilibrium constant. This constant is specific to a given temperature and reflects the balance between the dissolution and precipitation of the compound in a saturated solution. Once equilibrium is reached, no more net change in the concentration of any species is observed, although both the forward and reverse reactions continue to occur.
Ksp (Solubility Product Constant)
The solubility product constant, \(K_{sp}\), is a crucial concept in determining the solubility of a compound. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. The \(K_{sp}\) expression for a salt is derived from the concentration of the ions in the solution at saturation.

Each compound has a unique \(K_{sp}\) value. Smaller \(K_{sp}\) values indicate a lower solubility. In our example of the silver salt precipitates, \(\text{AgCl}\) with a \(K_{sp}\) of \(1.8 \times 10^{-10}\) has a higher solubility than \(\text{Ag}_2\text{CrO}_4\), which has a \(K_{sp}\) of \(1.2 \times 10^{-12}\). The smallest \(K_{sp}\) usually results in the fastest formation of a precipitate. When the ionic product exceeds the solubility product, precipitation occurs, thus establishing which compound precipitates first.

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Most popular questions from this chapter

Which of the following solutions is a buffer? (a) A solution made by mixing \(50 \mathrm{~mL}\), of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(250 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH},(\mathbf{b})\) A solution made by \(\mathrm{mix}\) ing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) nitric acid \(\left(\mathrm{HNO}_{3}\right),(\mathbf{c})\) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) potassium formate \((\mathrm{HCOOK})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KNO}_{3},\) (d) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\), and \(25 \mathrm{~mL}\). of \(0.200 \mathrm{MKOH}\).

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\), helps control the pH of physiological fluids, Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) per \(0.500 \mathrm{~L}\) of solution?

In nonaqueous solvents, it is possible to react HF to create \(\mathrm{H}_{2} \mathrm{~F}^{+} .\) Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, \((\mathbf{d})\) There is an acid in the nonaqueous medium that is a stronger acid than HE.

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{h}\) for B?
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