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Chapter 5: Problem 136

Equal volumes of the following \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\)solutions are mixed. In which of the solutions will precipitation occurs? \(\left[\right.\) Ksp of \(\left.\mathrm{CaF}_{2}=1.7 \times 10^{-10}\right]\) a. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{M} \mathrm{F}^{-}\) b. \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\) c. \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-2} \mathrm{M} \mathrm{F}^{-}\) d. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\)

Short Answer

Expert verified
Precipitation occurs in case d.

Step by step solution

01

Understand the equation for precipitation

The solution will precipitate calcium fluoride, \( \mathrm{CaF}_2 \), if the ionic product \( Q \) of the solution exceeds the solubility product constant \( Ks_p \). The ionic product \( Q \) for \( \mathrm{CaF}_2 \) is given by \( Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \). Precipitation occurs when \( Q > Ks_p \).
02

Calculate ion concentrations after mixing

Since equal volumes of solutions are mixed, the concentration of each ion is halved. For example, mixing \( 10^{-2} \text{ M} \) \( \mathrm{Ca}^{2+} \) and \( 10^{-5} \text{ M} \) \( \mathrm{F}^{-} \) will result in \( 5 \times 10^{-3} \text{ M} \) \( \mathrm{Ca}^{2+} \) and \( 5 \times 10^{-6} \text{ M} \) \( \mathrm{F}^{-} \).
03

Calculate ionic product Q for each case

For each pair of concentrations after mixing:- **Case a**: \( Q = (5 \times 10^{-3})(5 \times 10^{-6})^2 = 1.25 \times 10^{-13} \).- **Case b**: \( Q = (5 \times 10^{-4})(5 \times 10^{-4})^2 = 1.25 \times 10^{-10} \).- **Case c**: \( Q = (5 \times 10^{-5})(5 \times 10^{-3})^2 = 1.25 \times 10^{-10} \).- **Case d**: \( Q = (5 \times 10^{-3})(5 \times 10^{-4})^2 = 1.25 \times 10^{-9} \).
04

Compare each Q with Ksp

Given \( Ks_p = 1.7 \times 10^{-10} \):- **Case a**: \( Q = 1.25 \times 10^{-13} \), \( Q < Ks_p \), no precipitation.- **Case b**: \( Q = 1.25 \times 10^{-10} \), \( Q < Ks_p \), no precipitation.- **Case c**: \( Q = 1.25 \times 10^{-10} \), \( Q < Ks_p \), no precipitation.- **Case d**: \( Q = 1.25 \times 10^{-9} \), \( Q > Ks_p \), precipitation occurs.

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Key Concepts

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

Precipitation Reaction
Precipitation reactions are fascinating phenomena in chemistry. They occur when two soluble salts in aqueous solutions combine to form an insoluble solid. This solid is called a precipitate. A classic example is the mixing of calcium ions (\(\mathrm{Ca}^{2+}\)) and fluoride ions (\(\mathrm{F}^{-}\)) to form calcium fluoride (\(\mathrm{CaF}_2\)). During a precipitation reaction, if the ionic products formed exceed a specific value known as the solubility product constant (\(Ks_p\)), a precipitate forms. It acts like a threshold indicating the highest concentration of ions in solution before they "crash out" as a solid. Thus, precipitation reactions are crucial in understanding solubility and separation processes in chemistry.
Ionic Product
The ionic product (\(Q\)) is a vital concept in determining if precipitation will occur in a solution. It is calculated differently for various ionic compounds but follows a general rule: take the product of the concentrations of the ions raised to the power of their coefficients in the balanced chemical equation.For calcium fluoride, \(Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\). This equation signifies that \(Q\) is derived from the concentration of calcium ions and the square of the fluoride ions. By comparing \(Q\) with the solubility product constant (\(Ks_p\)), you can predict whether a precipitate will form:
  • If \(Q < Ks_p\), no precipitation occurs.
  • If \(Q > Ks_p\), precipitation occurs.
  • If \(Q = Ks_p\), the system is at equilibrium, and no further precipitate can form.
Calcium Fluoride Solubility
Calcium fluoride (\(\mathrm{CaF}_2\)) is a sparingly soluble compound, which means it dissolves only to a small extent in water. The solubility product constant (\(Ks_p\)) for calcium fluoride is a measure of its solubility. In this case, \(Ks_p = 1.7 \times 10^{-10}\).This value indicates the maximum product of the ion concentrations that can exist in a solution without forming a precipitate. Therefore, when you mix solutions containing calcium and fluoride ions:- These ions will dissolve to a degree dictated by \(Ks_p\).- If the concentrations after mixing lead to a \(Q\) value greater than \(Ks_p\), it suggests the formation of a solid precipitate as seen in the initial problem example for case d.
Chemical Equilibrium
Chemical equilibrium is a fundamental concept that describes a state where the rate of the forward reaction equals the rate of the backward reaction. In precipitation reactions, equilibrium governs the point at which a precipitate forms. When mixing \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\) ions, the solution eventually reaches a state where the dissolved ions continuously recombine to form more solid while the solid simultaneously dissolves back into ions.This balance is characterized by the solubility product constant (\(Ks_p\)):
  • If \(Q\) equals \(Ks_p\), the system is at equilibrium 鈥� no net change occurs in the precipitate amount.
  • If \(Q\) is greater than \(Ks_p\), the system shifts to reduce \(Q\), leading to precipitation.
Understanding chemical equilibrium helps in predicting the behavior of ionic reactions in solution and is vital for practical applications.

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Most popular questions from this chapter

\((\mathbf{A}): 0.33 \mathrm{M}\) solution of \(\mathrm{KCN}\) is more basic than \(0.33 \mathrm{M}\) solution of \(\mathrm{KF}\). (R): \(0.33 \mathrm{M}\) solution of \(\mathrm{KCN}\) is more basic than \(0.33 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOK}\).

The equilibrium constant \((\mathrm{Kp})\) equals \(3.40\) for the isomerization reaction: cis-2-butene \(\rightleftharpoons\) trans-2-butene If a flask initially contains \(0.250 \mathrm{~atm}\) of cis-2-butene and \(0.125\) atm of trans-2-butene, what is the equilibrium pressure of each gas? a. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.085 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.290 \mathrm{~atm}\) b. \(\mathrm{P}(\) cis-2-butene \()=0.058 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.230 \mathrm{~atm}\) c. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.028 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.156 \mathrm{~atm}\) d. \(\mathrm{P}\) (cis-2-butene) \(=0.034 \mathrm{~atm}\), \(\mathrm{P}(\) trans-2-butene \()=0.128 \mathrm{~atm}\)

\(\mathrm{Ag}^{+}+\mathrm{NH}_{3} \leftrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)^{+}\right]\) \(\mathrm{K}_{1}=3.5 \times 10^{-3}\) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+}+\mathrm{NH}_{3} \leftrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} ;\) \(\mathrm{K}_{2}=1.7 \times 10^{-3}\) Then the formation constant of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\)is a. \(6.08 \times 10^{-6}\) b. \(6.08 \times 10^{6}\) c. \(6.08 \times 10^{-9}\) d. None

\(\mathrm{N}_{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}, \mathrm{K}_{1}\) \(1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}, \mathrm{K}_{2}\) \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}, \mathrm{~K}_{3}\) \(\mathrm{NO} \rightleftharpoons 1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2}, \mathrm{~K}_{4}\) Correct relation between \(\mathrm{K}_{1}, \mathrm{~K}_{2}, \mathrm{~K}_{3}\) and \(\mathrm{K}_{4}\) is/are a. \(\sqrt{\left(K_{1} \times K_{4}\right)=1}\) b. \(\mathrm{K}_{1} \times \mathrm{K}_{3}=1\) c. \(\sqrt{\left(K_{1} \times K_{2}\right)=1}\) d. \(\sqrt{\mathrm{K}}_{2} \times \mathrm{K}_{2}=1\)

For the reaction $$ \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ the forward reaction at constant temperature is favoured by a. Increasing the volume of the container b. Introducing \(\mathrm{PCl}_{5}\) at constant volume c. Introducing an inert gas at constant volume d. Introducing \(\mathrm{Cl}_{2}(\mathrm{~g})\) at constant volume
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