18 -- Part 1: Solubility and Complex-Ion Equilibria
The molar solubility of a salt is determined by the value of the solubility product.
Discuss how the addition or presence of a common ion affects the solubility of a small amount of salt.
Explain how fractional precipitation can be used to separate ions.
Discuss how the pH of a solution may affect the solubility of a salt.
The formation constant, K, is used to determine the concentrations of ion in solution.
Explain how to use precipitation in a cavern in Liguria, Italy.
stalagmites are formed from calcium salts deposited underground.
The formation of limestone caverns is one of the natural phenomena caused by the dissolution and precipitation of limestone.
There is a need to combine ideas about acid-base equilibria from Chapters 16 and 17 with ideas about the new types of equilibria to be introduced in this chapter.
There are additional topics discussed in this chapter.
Some dissolved calcium sulfate can be found in the ground that comes into contact with gypsum.
The water can't be used for cooling power plants because of the calcium sulfate in the water.
Appendix D has a more extensive listing of Ksp values.
CaF2 is one of the substances used in a fluoride treatment.
Cu3(AsO4)2 is used as an insecticidal and fungicide.
The equation for the equilibrium is for one mole of the solute.
The slightly soluble solute is understood by the coefficients 1 and 1.
The coefficients are used to balance the equation.
The powers to which the ion concentrations are raised are established by the coefficients.
The Ksp expression does not show the solid.
Ag3PO4 is used in photographic emulsions.
A handbook lists Ksp as 1 * 10-7 for calcium hydrogenphosphate, a substance used in dentifrices and as an animal feed supplement.
A large excess of MgF2(s) is maintained in contact with pure water to produce a saturated solution of MgF2.
The calculations involving Ksp are found in the human kidneys.
In practice example 18-3B, additional conversions are required to get solubility in units other than moles per liter.
CaSO4 is listed as 0.20 g CaSO4> 100 mL in a handbook.
The Ksp expression can be replaced with 4, which we can do.
The first step is to convert the mass of CaSO4 into a liquid.
The inverse of the molar mass of CaSO4 is used to replace 100 mL.
The value in Table 18.1 is different from the Ksp result determined here.
There is a discrepancy because ion activities need to be taken into account.
A handbook shows the AgOCN's solubility at 20 degrees.
A handbook says the Li3PO4 is 100 mL soln at 18 degC.
PbI2 appeared in the solution.
The Ksp expression must be satisfied by these concentrations.
PbI has a molar solubility in water of 1.2.
We have to account for the correct number of moles in each species.
We had 2 moles of iodide ion for every mole of lead.
The Ksp of Cu3(AsO4)2 at 25 degC is 7.6.
Ksp is 1.1
The solute with the largest Ksp value will have the greatest molar solubility.
The Ksp is more important than the AgBr.
Even though Ag2CrO4's product constant is smaller, it is still moresoluble than Cnri/Science Source AgCl 1Ksp.
This soft tissue will show up when X-rayed if CONCEPT ASSESSMENT is done.
Our first encounter with the common-ion effect was in Chapter 17.
An equilibrium mixture responds to a forced increase in the concentration of one of its reactants by shifting in the direction in which that reactant is consumed.
Adding the common ion, I-, causes the reverse reaction to be favored, leading to a new equilibrium.
The equilibrium changes to form more PbI.
The addition of the common ion causes the equilibrium of a slightly soluble ionic compound to shift towards the undissolved compound.
The compound's solubility is reduced.
In the presence of a second solute, the ionic compound's solubility is lowered.
The solubility of PbI2 in the presence of 0.10 M I- is 2000 times less than in pure water.
The effect of added Pb2+ in reducing the solubility of PbI2 is not as striking as that of I-, but it is significant nonetheless.
Excess undissolved solute has been removed from a clear saturated solution of lead.
The Carey B. V solute is reduced by a common ion.
Think of a saturated solution of PbI2, but instead of using pure water as the solvent, we will use 0.10 M KI(aq).
There are two additional concentrations in this solution.
The Ksp relationship needs to be happy.
The molar solubility of PbI2 is 7.1.
2(7.1 * 10-7) is much smaller than 0.10.
A typical error in such problems is to double the common-ion concentration, that is, to write instead of 3I-4.
There is no relationship between the dissolution of PbI2 and the establishment of 3I-4, which requires a factor of 2.
Is it possible to write Ksp solute dissolved in the expressions for solvent?
There are no interac NaCl, KNO3 or NaOH.
In an ideal solution where high concentrations, activities and concentrations are not equal, is where ionic solutions are moderate.
The activity is roughly 25% of the concentrations.
The interactions increase in molarity.
The sim more solute is dissolved if we can't use molarities in place of activities.
The common-ion effect is 0 to 0.10 M added salt.
Concentration of salt, M limited to insoluble solutes, and ion molarities are used in place of activities.
The Ksp concept has several limitations, which are discussed in the following sections.
Diverse ions tend to increase in size.
Interionic attractions become more important as the total ionic concentration increases.
The numerical value of a Ksp will vary depending on the ionic atmosphere.
The problem of the salt effect is avoided by most Ksp values being based on activities.
We have assumed that the dissolved solute appears in solution as separated cations and anions.
This assumption is often not valid.
Some of the solute might enter solution in a molecule.
An ion pair is two oppositely charged ion that are held together.
If a solution contains cations and anions of a solute, the concentrations of the dissociated ion are reduced from the stoi solution.
Although the measured solubility of MgF surrounding the ion pair form 2 is what is referred to as a solvent, we cannot assume that 3Mg2+4 is 4 * 10-3 M, and cage.
The true solubility of the solute is greater than expected on the basis of Ksp if additional solute is present.
The reaction between a solid solute and its ion in a solution is only one part of the process.
We can generally ignore the self-ionization of water.
Other equilibrium processes include reactions between solution species.
There are two possibilities, acid-base reactions and complex-ion formation.
If we don't take into account other equilibrium processes that occur simultaneously with solution equilibrium, we may be in error.
In this chapter, we have seen the dissolution of PbI2(s) several times.
The dissolution process is more complex than we have shown.
The result was 2.3 * 10-4.
This value is 25 times larger than the value listed in Table 18.1.
There are conflicting results for CaSO4.
The Ksp value listed in Table 18.1 is based on ion activities, while the Ksp value calculated from the experiment is based on ion concentrations.
The case of CaSO4 suggests that some of our results, although within a factor of 10 or 100, may not be highly accurate.
The order-of-magnitude results allow us to make some predictions and apply the Ksp concept in useful ways.
Criteria for Precipitation and Its Completeness Silver iodide is a light-sensitive compound used in photographic film and cloud seeding to produce rain.
It has the same form as an equilibrium constant expression but uses initial concentrations.
A net change should occur to the left because Qsp 7 Ksp are already higher than they would be in a saturated solution.
Excess AgI should come from the solution.
There wouldn't be a form of precipitation from such a solution.
We need to compare the ion product with Ksp to determine whether the ion product will combine to form a precipitate.
If Qsp is Ksp, a solution is just saturated.
If the amount of solute remaining in solution is small, precipitation is complete.
The criteria for precipitation must be applied.
If more than 99% of a particular ion has precipitated, the precipitation is complete.
The completeness of the precipitation will be determined by comparing this remaining 3Mg2+4 to the initial 3Mg2+4.
Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO3).
We need to compare the initial concentrations with the Ksp for PbI2.
3Pb2+4 to 0.100 L of solution is negligible and so we can simply use 0.010 M.
The value of [Pb2+]: 0.010 M * (0.1000 L/0.1002 L) is the same as before.
Three drops of 0.20 M KI are added to a 0.010 M solution of AgNO3.
The first step in a commercial process in which magnesium is obtained from seawater involves precipitating magnesium.
We compare the Qsp of the solution with the known Ksp for Mg(OH)2 to see if precipitation will occur.
As long as the ion product is greater than Ksp, precipitation of Mg(OH)2(s) will continue.
The original amount of magnesium remains after precipitation.
There is no question that precipitation will occur.
As a result of the precipitation reaction, 3Mg2+4 is reduced from 0.059 M to 4.5 M.
We conclude that precipitation is essentially complete because less than 1% of the Mg2+ remains.
The calculation of [Mg2+] is straightforward because [OH- ] is maintained at a constant value.
You can verify this result yourself.
The value of Ksp, the initial concentration of the target ion, and the concentration of the common ion are some of the factors that can be used to determine whether the target ion is completely removed from solution in a precipitation.
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