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Chapter 18: Problem 91

Briefly describe each of the following ideas, methods, or phenomena: (a) common-ion effect in solubility equilibrium; (b) fractional precipitation; (c) ion-pair formation; (d) qualitative cation analysis.

Short Answer

Expert verified
The common-ion effect refers to the decrease in a salt's solubility due to a common ion's presence. Fractional precipitation separates ions in a solution based on solubility differences in their salts. Ion-pair formation involves a cation and anion forming a neutral compound in a solution. Qualitative cation analysis detects specific cations' presence in a sample based on their reaction qualities.

Step by step solution

01

Part (a) - Common-ion effect in solubility equilibrium

The common-ion effect in solubility equilibrium refers to the decrease in the solubility of a salt when a common ion is added to the solution. It can be explained using Le Chatelier's Principle, which states that the reaction will adjust so as to minimize the effect of the added ion.
02

Part (b) - Fractional precipitation

Fractional precipitation is a method used to separate ions in a solution using the differences in solubilities of their salts. The solution is mixed with a reagent that forms insoluble salts with the target ions. The target ions are then separated by precipitating one ion at a time by carefully adjusting the concentration of the reagent.
03

Part (c) - Ion-pair formation

Ion-pair formation refers to the process where a cation and anion in a solution come together to form a neutral compound. This is a common occurrence in high ionic strength solutions. Though they are neutral, ion pairs can have significant effects on the thermodynamics and kinetics of reactions.
04

Part (d) - Qualitative cation analysis

Qualitative cation analysis is a method used to detect presence of specific cations in a given sample. It typically involves responding to color changes, formation of precipitates or bubble formation and using those reactions to identify the presence of specific cations.

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Most popular questions from this chapter

A 725 mL sample of a saturated aqueous solution of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4},\) at \(95^{\circ} \mathrm{C}\) is cooled to \(13^{\circ} \mathrm{C}\). How many milligrams of calcium oxalate will precipitate? For \(\mathrm{CaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-8}\) at \(95^{\circ} \mathrm{C}\) and \(2.7 \times 10^{-9}\) at \(13^{\circ} \mathrm{C}\).

Will the following precipitates form under the given conditions? (a) \(\mathrm{PbI}_{2}(\mathrm{s}),\) from a solution that is \(1.05 \times 10^{-3} \mathrm{M} \mathrm{HI}\), \(1.05 \times 10^{-3} \mathrm{M} \mathrm{NaI},\) and \(1.1 \times 10^{-3} \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). (b) \(\operatorname{Mg}(\mathrm{OH})_{2}(\mathrm{s}),\) from \(2.50 \mathrm{L}\) of \(0.0150 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) to which is added 1 drop \((0.05 \mathrm{mL})\) of \(6.00 \mathrm{M} \mathrm{NH}_{3}\). (c) \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) from a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Al}^{3+}, 0.010 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH},\) and \(0.010 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\).

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-11}\right)\) in (a) pure water; (b) \(0.0862 \mathrm{M}\) \(\mathrm{MgCl}_{2} ;\) (c) \(0.0355 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\).

Fluoridated drinking water contains about 1 part per million (ppm) of \(\mathrm{F}^{-}\). Is \(\mathrm{CaF}_{2}\) sufficiently soluble in water to be used as the source of fluoride ion for the fluoridation of drinking water? Explain. [Hint: Think of 1 ppm as signifying \(1 \mathrm{g} \mathrm{F}^{-}\) per \(10^{6} \mathrm{g}\) solution.

An aqueous solution that \(2.00 \mathrm{M}\) in \(\mathrm{AgNO}_{3}\) is slowly added from a buret to an aqueous solution that is \(0.0100 \mathrm{M}\) in \(\mathrm{Cl}^{-}\) and \(0.250 \mathrm{M}\) in \(\mathrm{I}^{-}\). (a) Which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is the first to precipitate? (b) When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c) Is the separation of \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\) feasible by fractional precipitation in this solution?
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