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Chemistry 12Unit 3 – Solubility Equilibrium

1. Introduction to solubility

• Solution are examples of homogeneous mixtures because they are uniform throughout. The two components of a mixture are the solute(substance in lesser concentration)and the solvent (substance in greater concentration).

Unit 3 - Solubility Equilibrium 2

• Substances that dissolve in water can either be :

(1) Electrolytes

• A substance that dissolves to give an electrically conducting solution containing ions.

Na3PO4(s) → 3Na+(aq) + PO43- (aq)

KCl(g) → K+(aq) + Cl-(aq)

• The resulting solution is called an ionic solution.

• The constituents, called ionic compounds, are made up of a metaland a non-metal:

FeCl3(s) → Fe3+ (aq) + 3Cl- (aq)

Na2SO4(s) → 2Na+(aq) + SO42- (aq)


(2) Non-electrolytes

• A substance that dissolves to give a non-conducting solution containing only neutral molecules.

CH3OH(g) → CH3OH(aq)

Br2(I) → Br2(aq)

• The resulting solution is called a molecular solution.

• Molecular solution contains molecular compounds formed between non-metals.

C2H5OH(I) → C2H5OH(aq)

CIO2(I) → CIO2(aq)


• The solubility of a substance is the maximum amount of the substance that can dissolve in a given amount of solvent at a given temperature. It is usually reported as molar solubility in a unit of mol/L, or M.

• When no more solid can be dissolved into a solution, it is said to be saturated. There are two conditions associated with saturation:

• (1) Some undissolved material is present in the solution.

• (2) There exists an equilibrium between the dissolved ions and the undissolved solid.

• In simple words, dissolution of a solid in a solvent is an equilibriumprocess!


• How to tell if a solution is saturated?


• Since dissolving a substance into a solution is an equilibrium process, we can write the corresponding equation in form of an equilibrium between the substance and its aqueous ions.

• For instance, dissolving silver sulfate in water can be described by the following:

Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)


• How can we interpret the equilibrium?

• (1) Dissolving reaction: Solid Ag2SO4 dissociating into ions.

Ag2SO4(s) → 2Ag+(aq) + SO42-(aq)

• (2) Crystallization reaction: Ag+ and SO42- ions come together and

form Ag2SO4.

2Ag+(aq) + SO42-(aq) → Ag2SO4(s)


• From kinetics perspective, dissolution can be understood as:

• Initially, there are few ions in solution and the dissolving reaction predominates. The crystallization reaction speeds up as ion concentration increases.

• Eventually, the rate of the dissolving reaction equals the rate of the crystallization reaction and equilibrium is reached.


2. Calculating solubility and ion concentration

• Solubility is usually expressed as molar solubility with any of the following units

(1) mol/L

(2) g/L

(3) g/100mL of solution

• It is necessary to be able to convert between units of solubility.

• To find solubility experimentally , it is necessary to find the mass of solute required to make a known volume of a saturated solution.


• Example: If the solubility of AgI is 3.45 g/100 mL of solution, what is the solubility in terms of g/L and mol/L?


• Example: If 0.750L of a saturated AgCl solution contains 2.50g AgCl , what is the molar solubility of AgCl?


• Example: If the molar solubility of PbI2 is 1.37 x 10-3 M, how much grams of PbI2 will dissolve in 450 mL?


• When an ionic compound dissolves, it dissolves into ions. For example,

Na3PO4(s) → 3Na+(aq) + PO43-(aq)

Ba(NO3)2(s) → Ba2+(aq) + 2NO3

-(aq)

• It is important to be able to write balanced dissociation equation as well as calculate the concentrations of each of the ions in solution.

• When writing dissociation equations, remember that only two kinds of ions are formed and often they will be polyatomic ions.


• Example: What are the concentrations of ions in 0.25 M Na3PO4(aq)?


• Example: What are the concentrations of ions when 150 mL of 0.250 M Na3PO4(aq) is mixed with 250 mL of 0.750 M NaCl(aq)?


3. Predicting the solubility of salts

• Recall that dissolution is an equilibrium process comprising a dissolving step and a crystallization step. These two steps may show a huge difference in terms of favor (either the solution form or the solid form is preferred), but they both occur naturally. In other words, nothing is truly insoluble in water.

• Despite this, we can still classify substances as soluble or slightly soluble in water with regard to a reference solubility at 25°C.


Types of substances Molar solubility

Soluble > 0.1 M

Slightly soluble < 0.1 M

• Experimentally chemists have studied the solubility of many inorganic compounds, and observed a number of general rules.

• (1) All Group 1 ionic compounds are soluble in water regardless of the type of the anion.

• Group 1 ions refer to Li+, Na+, K+, Rb+, Cs+ and Fr+

• (2) Ammonium salts are all soluble in water.

• Ammonium ion (NH4+) in many ways behaves like Group 1 ions

• (3) All nitrates (NO3-) are soluble in water regardless of the type of

cation.

• (4) Ionic compounds containing proton (H+) are soluble in water.


• How do we use the Solubility Table?

• It tells us that all alkali phosphates, such as Li3PO4, alkali carbonates, such as K2CO3, and alkali sulphites, such as Rb2SO3, are soluble, but other compounds, such as BaSO3, are not.


• Hence, this table gives us ideas whether precipitate is expected when two solutions are mixed.

• For example: Which of the following compounds have low solubility?

Na3PO4, MgSO3, ZnCO3, (NH4)2CO3


They are soluble They have low solubility

• Example: Will a precipitate form when equal volumes of 0.1 M CaSand 0.1 M Na2SO4 are mixed?

• Solution: We first write the equation for this process

CaS + Na2SO4 → CaSO4 + Na2S

• From the solubility table, we know that Na2S is soluble but CaSO4 is not. So, we expect a precipitate.

• Is it true?

• Check the concentration: The concentration after mixing is

• Since the concentration is only 0.05 M, we will not see precipitate indeed.


𝑀2 =𝑀1𝑉1

𝑉2=

(0.1)𝑉1

2𝑉1= 0.05

• Example: What compound could precipitate SO42-(aq) from a

solution?

• Solution: From the solubility table, we find many cations which form insoluble salts with sulfate, such as Ba2+.

• However, we can’t add Ba2+ on its own. We should choose a soluble salt such as Ba(NO3)2.


• In practice, when we are to precipitate a particular ion from a solution, we use a compound containing the ion that can form insoluble salt with the target ion.

• In order to prevent further complications, we choose the desired compound using the following guideline:


Ion to be added Desired compound

Cation Nitrate salt

Anion Sodium salt

4. Selective precipitation• The idea of precipitation is useful in chemical analysis if we want to

identify the ions present in a mixture.

• If two cations form salts with the same anion with substantially different solubility, then the one with a lower solubility would be precipitated out, leaving the other one in the solution. This process is called selective precipitation.

• Selective precipitation is a very handy and powerful strategy in chemical qualitative analysis.


• For instance, there is a solution containing Sr2+ and Mg2+. How could we isolate them?

• From the solubility table, we realize that


• Based on this information, we can design an experiment:

• (1) Add NaOH into the mixture

• Mg2+ will be precipitated out as Mg(OH)2.

• (2) Add Na2SO4 into the mixture

• Sr2+ will be precipitated out as SrSO4.


Mg2+

Sr2+Sr2+

NaOH Na2SO4

• Example: A solution contains either Ag+ or Sr2+ ions. Devise a method of determining which ion is present in the solution.

• Solution: Consult the following table

• We can try Cl-, S2- or OH-.


• Example: How can we separate Ag+, Ba2+, and Ni2+ from a solution mixture?

• Solution: Consider the solubility table:

• Which reagents should be used? And the sequence?


5. The solubility product

• When a substance is dissolved in water, an equilibrium is formed between the solid and its ions. For example:

PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

• Remember that solubility equilibrium is dynamic because both the dissolving and crystallization processes continue to occur even though equilibrium has been established.

• When equilibrium is reached, the solution is said to be saturated.


• Since dissolution is an equilibrium process, it can be described by means of equilibrium constant. By definition, the equilibrium constant expression, called solubility product, is given by

• PbCl2 is a solid; therefore it is replaced apparently by a constant “1” and the expression becomes

• Solubility product is an ion product; it consists only of ionic species. This constant has no unit and is dependent upon temperature!


Ksp =Pb2+ Cl− 2

PbCl2

Ksp = Pb2+ Cl− 2

• To obtain a correct Ksp expression for a dissolution process, it is important to know how a chemical equation can be written accurately.

• An chemical equation can be written in 3 different ways. For instance, for the reaction of Ca(NO3)2 and Na3PO4:

(1) Formula equation

3Ca(NO3)2(aq) + 2Na3PO4(aq) → 6NaNO3(aq) + Ca3(PO4)2(s)

(2) Complete ionic equation

3Ca2+(aq) + 6NO3-(aq) + 6Na+(aq) + 2PO4

3-(aq)

→ 6Na+(aq) + 6NO3-(aq) + Ca3(PO4)2(s)


(3) Net ionic equation

3Ca2+(aq) + 2PO43-(aq) → Ca3(PO4)2(s)

• Notes:

(a) In a net ionic equation, all spectator ions, which do not take part in the reaction and appear on both sides of the equation, are removed.

(b) Net ionic equation describes the formation of precipitate only. To construct it, just look for the species involved in the precipitate.


• Example: Write the Ksp expression for the following equilibria.

(a) BaCrO4(s) ⇌ Ba2+(aq) + CrO42-(aq)

(b) Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq)


• Just like chemical equilibrium in general, solubility equilibrium can be reached from either the reactants or the products.

• Consider the process: PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

From the reactant side:

• PbCl2(s) is simply dissolved in water and dissociates to the corresponding ions in stoichiometry. Therefore,

• This ratio remains the same regardless of time.


Cl− = 2 × Pb2+

From the product side:

• The ratio of Pb2+ and Cl− is no longer 1:2. It depends on the solutions that are mixed together.

• For example: When 25.0 mL of 0.025 M Pb(NO3)2 solution is mixed with 30.0 mL of 0.010 M AlCl3 solution,


Pb2+ =(0.025)(0.025)

(0.025 + 0.030)×

1 Pb2+

1 Pb NO3 2= 0.0114 M

Cl− =(0.030)(0.010)

(0.030 + 0.025)×

3 Cl−

1 AlCl3= 0.0164 M

6. Calculations involving Ksp

• The calculations involving Ksp are easier than those in Chapter 2; only the products have to be considered. The reactant is ignored since it is in solid state.

• In general, there are two types of calculations involving Ksp:

(1) Determine the value of Ksp from solubility data

(2) Determine the value of solubility from Ksp


• Example: 100 mL of a saturated PbI2 solution was found to contain 5.23 × 10-2 g PbI2. Calculate Ksp.

• Solution: We first find the molar solubility of PbI2:

• Hence,

• And


Mole of PbI2 =5.93 × 10−2

461= 1.29 × 10−4 mol

PbI2 =1.29 × 10−4

0.100= 1.29 × 10−3 M

Pb2+ = 1.29 × 10−3 M I− = 2 1.29 × 10−3 = 2.58 × 10−3 M

Ksp = Pb2+ I− 2 = 1.29 × 10−3 2.58 × 10−3 2 = 8.59 × 10−9

• Example: What is the molar solubility of PbCl2 if the Ksp of PbCl2 is 1.8 × 10-4?

• Solution: Note that PbCl2 ⇌ Pb2+ + 2Cl-

• Hence,

• Assume PbCl2 = Pb2+ = 𝑥, then Cl− = 2𝑥. Therefore,


Ksp = Pb2+ Cl− 2

1.8 × 10−4 = 𝑥 2𝑥 2 = 4𝑥3

𝑥 =3 1.8 × 10−4

4= 0.0356 M

• Example: What is the concentration of [Ag+] in a saturated solution of Ag2CO3?

• Solution: The Ksp of Ag2CO3 at 25°C is 8.5 × 10-12.

• Let 𝑥 be the molar solubility of Ag2CO3. Hence,

• And

• On solving, 𝑥 = 1.29 × 10−4 M, and


Ag+ = 2𝑥 CO32− = 𝑥

Ksp = Ag+ 2 CO32−

8.5 × 10−12 = 2𝑥 2𝑥 = 4𝑥3

Ag+ = 2 1.29 × 10−4 = 2.57 × 10−4 M

7. Trial ionic products

• When two solutions containing the ions that will form insoluble salt are mixed, precipitate is expected to form. However, in reality, this is not always the case.

• In fact, whether a precipitate is seen or not depends on the actual concentrations of the ions in the solution after mixing.

• Because of this, a new term, called trial ionic product, is defined. For a solubility equilibrium:

• Trial ionic product (TIP):


𝐴𝑎𝐵𝑏(𝑠)⇌ 𝑎𝐴+ + 𝑏𝐵−

TIP = A+ 𝑎 B− 𝑏

• TIP is a quantity that tells us if the solubility equilibrium is shifting to the left, to the right, or has been established.

• Case 1: TIP = 𝐾𝑠𝑝

• In this case, the equilibrium is established and no more solid can be dissolved.

• Case 2: TIP > 𝐾𝑠𝑝

• In this case, there are too many ions in the solution. The excess ions will recombine to form precipitate.

• Case 3: TIP < 𝐾𝑠𝑝

• In this case, there are too few ions in the solution. Therefore, more solid will dissolve.


• These will imply the followings:

(1) If two solutions are mixed such that 𝑇𝐼𝑃 > 𝐾𝑠𝑝 of the insoluble

salt, then the solution is over-saturated and precipitate will be formed.

(2) If two solutions are mixed such that 𝑇𝐼𝑃 = 𝐾𝑠𝑝 of the insoluble

salt, then the solution is saturated and no precipitate will be formed.

(3) If two solutions are mixed such that 𝑇𝐼𝑃 < 𝐾𝑠𝑝 of the insoluble

salt, then the solution is under-saturated and no precipitate will be formed.


• Example: Will a precipitate form when 23 mL of 0.020 M Na2CO3 is added to 12 mL of 0.010 M MgCl2?

• Solution: The Ksp of MgCO3 is 6.8 × 10-6.

• The concentrations of Mg2+ and CO32- after mixing are:

• Therefore,

• Since 𝑇𝐼𝑃 > 𝐾𝑠𝑝, precipitate is formed.Unit 3 - Solubility Equilibrium 46

Mg2+ =0.012 × 0.010

0.012 + 0.023= 0.0034 M

CO32− =

0.023 × 0.020

0.012 + 0.023= 0.013 M

TIP = Mg2+ CO32− = 0.034 0.013 = 4.4 × 10−5

• Example: If the Ksp for PbCl2 is 1.8 × 10-4, will a precipitate form when 200.0 mL of 0.015 M NaCl is mixed with 100.0 mL of 0.060 M Pb(NO3)2?


• Example: What mass of solid sodium sulfate is required to start precipitation in 500 mL of 0.030 M calcium chloride at 25°C? Assume the volume remains the same.

• Solution: We first write the equilibrium equation

• Note that all Ca2+ ions come from calcium chloride; therefore, its concentration is

• The Ksp value for CaSO4 at 25°C is 7.1 × 10-5. Hence,


CaSO4 s ⇌ Ca2+ aq + SO42−(aq)

Ca2+ = 0.030 M

Ksp = Ca2+ SO42−

• and

which gives

• Therefore, the mass of Na2SO4 required is:


7.1 × 10−5 = 0.030 SO42−

SO42− =

7.1 × 10−5

0.030= 2.4 × 10−3 M

m = 2.4 × 10−3 0.500 142.1 = 0.17 g

• Example: What is the maximum [SO42-] that can exist in a saturated

solution of CaCO3?

• Solution: We have to first find out the saturated concentration of Ca2+

ions in the solution. Given the Ksp of CaCO3 is 5.0 × 10-9, we have

• Hence

• From the solubility table, we know that the Ksp for CaSO4 is 7.1 × 10-5

at the same temperature. That means,

and


Ksp = Ca2+ CO32− = 𝑥2 = 5.0 × 10−9

Ca2+ = 𝑥 = 7.1 × 10−5 M

Ksp = Ca2+ SO42− = 7.1 × 10−5 SO4

2− = 7.1 × 10−5

SO42− = 1.0 M

8. The common ion effect

• Recall that dissolution is an equilibrium process between a heterogeneous mixture of a solid and its solution. The extent to which a solid dissolves depends on the temperature. (Just like other types of equilibrium processes.)

• However, we have not yet looked at the influence brought forth by the solvent. Until now, we only assumed the solvent to be a purewater which contains no contaminants or any ions.

• How is the degree of dissolution of a solid affected by the presence of other ions in the solvent?


• Consider a saturated PbCl2 solution:

• The forward reaction is a dissolving process:

• If any action to this system causes the equilibrium to shift to the right, the rate of PbCl2 solid dissolving into the solution will increase.

• This will give rise to an enhanced solubility of PbCl2.


PbCl2 s ⇌ Pb2+ aq + 2Cl−(aq)

PbCl2 s → Pb2+ aq + 2Cl−(aq)

• On the other hand, the reverse reaction is a recrystallization process:

• If any action to the system causes the equilibrium to shift to the left, then the rate of reforming the PbCl2 solid will increase.

• This will give rise to a reduced solubility of PbCl2.

• What kinds of action should we do in order to achieve these shifts?


PbCl2 s ← Pb2+ aq + 2Cl−(aq)

(1) Shift to the right

• Due to the Le Châtelier’s principle, if either Pb2+ or Cl- ions is removed, the instantaneous TIP will become smaller than the Ksp, thereby driving the PbCl2 solid to dissolve more.

• These ions can be removed by introducing appropriate counterions which form a more insoluble precipitate than PbCl2.

• For example, Pb2+ can be precipitated out from a saturated PbCl2solution if NaBr is added, in which insoluble PbBr2 is formed.


• Similarly, AgNO3 can be added to remove Cl- because AgCl has a smaller Ksp than PbCl2:


(2) Shift to the left

• Since PbCl2 is a solid, it does not appear in the Ksp expression. Removing the solid does not affect the equilibrium and thus the solubility of PbCl2.

• Instead, additional Pb2+ or Cl- ions can be added to a saturated solution to increase the instantaneous TIP. This drives the equilibrium to shift to the left, forming more PbCl2. As a result, the solubility of PbCl2 is reduced.

• This can be achieved by adding soluble compounds containing Pb2+ or Cl-.


• Example: What is the solubility of Mg(OH)2 in (a) water? (b) 0.10 M MgCl2?

• Solution: The Ksp of Mg(OH)2 is 5.6 × 10-12.

• (a) In water:

• Therefore, the solubility of Mg(OH)2 is:


Ksp = Mg2+ OH− 2 = 𝑥 2𝑥 2 = 4𝑥3 = 5.6 × 10−12

Mg OH 2 = Mg2+ = 𝑥 = 1.1 × 10−4 M

• (b) In 0.10 M MgCl2:

• The concentration of Mg2+ at the beginning is 0.10 M. Hence,

• Assuming 𝑥 is very small compared to 0.10, we have

• Therefore,

• It yields


Ksp = Mg2+ OH− 2 = 0.10 + 𝑥 2𝑥 2 = 5.6 × 10−12

0.10 + 𝑥 ≈ 0.10

0.10 4𝑥2 = 0.4𝑥2 = 5.6 × 10−12

Mg(OH)2 = Mg2+ = 𝑥 = 3.7 × 10−6 M

• Example: Calculate the molar solubility of silver iodate in 0.12 M sodium iodate.


Practice Questions (from previous provincials)

• Question 1: A 30.00 mL sample of a saturated solution of Ag2SO4 was heated in an evaporating dish until all the water was evaporated. The following data were recorded:

Calculate the Ksp value of Ag2SO4.


Volume of solution 30.00 mL

Mass of evaporating dish 32.125 g

Mass of evaporating dish and solid Ag2SO4 32.260 g

• Question 2: Calculate the solubility of SrSO4 in grams per liter.


• Question 3: A 100 mL saturated solution of FeF2 contains 0.0787 g of solute. Determine the Fe2+ and F− in the solution.


• Question 4:

• Calculate the Cl− in the water sample.


• Question 5: Calculate the mass of NaI necessary to begin precipitation of Cu+ from a 250.0 mL sample of 0.010 M CuNO3.


• Question 6: Determine, with calculations, whether a precipitate will form when 15.0 mL of 0.050 M Pb(NO3)2 is added to 35.0 mL of 0.085 M NaCl.


• Question 7: Calculate the maximum mass of BaCl2 that can be added to 250 mL of 0.50 M Pb(NO3)2 without forming a precipitate of PbCl2.


• Question 8: Write the chemical equation, complete ionic equation, and net ionic equation for the reaction between NaBrO3 and AgNO3.
