Chapter 17 Lecture- Buffers & Ksp

Chapter 17- Aqueous Equilibria

Section 17.1 The Common Ion Effect

The Common-Ion Effect

• Consider a solution of acetic acid:

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)


• Consider a solution of acetic acid:

• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)


“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”


Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Ka for HF is 6.8 × 10−4.


Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Ka for HF is 6.8 × 10−4.

[H3O+] [F−][HF]

Ka= = 6.8 × 10-4


Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F−], M

Initially 0.20 0.10 0

Change −x +x +xAt Equilibrium 0.20 − x ≈ 0.20 0.10 + x ≈ 0.10 x

HF(aq) + H2O(l) H3O+(aq) + F−(aq)


= x

1.4 × 10−3 = x

(0.10) (x)(0.20)6.8 × 10−4 =

(0.20) (6.8 × 10−4)(0.10)


• Therefore, [F−] = x = 1.4 × 10−3

[H3O+] = 0.10 + x = 0.10 + 1.4 × 10−3 = 0.10 M


• Therefore, [F−] = x = 1.4 × 10−3

[H3O+] = 0.10 + x = 1.01 + 1.4 × 10−3 = 0.10 M

• So, pH = −log (0.10) pH = 1.00


Section 17.2 Buffers

Bellwork

a) [NH4+] = 0.10 M

[Cl–] = 0.10 M [NH3] = 0.12 M

Bellwork

a) [NH4+] = 0.10 M

[Cl–] = 0.10 M [NH3] = 0.12 M b) Cl–

Bellwork

a) [NH4+] = 0.10 M

[Cl–] = 0.10 M [NH3] = 0.12 M b) Cl–c)NH3(aq) + H2O(l) NH4

+(aq) + OH–

(aq)

Bellwork

Buffers:

Buffers:

• Solutions of a weak conjugate acid-base pair.

Buffers:

• Solutions of a weak conjugate acid-base pair.

• They are particularly resistant to pH changes, even when strong acid or base is added.

HNO3 and NO3– will not work as a buffer

because NO3– is a spectator ion.

Buffers

If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Buffers

If acid is added, the F− reacts to form HF and water.

The NaOH reacts with HC2H3O2 converting some of it into C2H3O2

–.

Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

HA + H2O H3O+ + A−

Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

[H3O+] [A−][HA]

Ka =

HA + H2O H3O+ + A−

Buffer Calculations

Rearranging slightly, this becomes

[A−][HA]Ka = [H3O+]

Buffer Calculations


[A−][HA]Ka = [H3O+]

Taking the negative log of both sides, we get

Buffer Calculations


[A−][HA]Ka = [H3O+]

Taking the negative log of both sides, we get

[A−][HA]−log Ka = −log [H3O+] + −log

pKapH acid

base

Buffer Calculations• So

pKa = pH − log [base][acid]



• Rearranging, this becomes

pH = pKa + log [base][acid]



• Rearranging, this becomes


• This is the Henderson–Hasselbalch equation.

Henderson–Hasselbalch Equation

What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 × 10−4.

Henderson–Hasselbalch Equation


pH = −log (1.4 × 10−4) + log(0.10)(0.12)

pH = 3.85 + (−0.08)

pH = 3.77

PRACTICE EXERCISECalculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.)

PRACTICE EXERCISECalculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.)

Answer: 4.42

How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume)


The major species in the solution will be NH4+, Cl–, and NH3. Of these, the Cl–

ion is a spectator (it is the conjugate base of a strong acid).




and so




and so

Because Kb is small and the common ion NH4+ is present, the equilibrium

concentration of NH3 will essentially equal its initial concentration:

Comment: Because NH4+ and NH3 are a conjugate acid-

base pair, we could use the Henderson–Hasselbalch equation to solve this problem. To do so requires first using Equation 16.41 to calculate pKa for NH4

+ from the value of pKb for NH3. Try this approach to convince yourself that you can use the Henderson–Hasselbalch equation for buffers for which you are given Kb for the conjugate base rather than Ka for the conjugate acid.

PRACTICE EXERCISECalculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.00.

Answer: 0.13 M

PRACTICE EXERCISECalculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.00.

pH Range

• The pH range is the range of pH values over which a buffer system works effectively.

pH Range

• The pH range is the range of pH values over which a buffer system works effectively.

• It is best to choose an acid with a pKa close to the desired pH.

pH = 4.74, this is pKa of acetic acid

When Strong Acids or Bases Are Added to a Buffer…

When Strong Acids or Bases Are Added to a Buffer…

…it is safe to assume that all of the strong acid or base is consumed in the reaction.

Addition of Strong Acid or Base to a Buffer

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

Addition of Strong Acid or Base to a Buffer

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

Calculating pH Changes in Buffers

Before the reaction, since mol HC2H3O2 = mol C2H3O2

−

pH = pKa = −log (1.8 × 10−5) = 4.74

Calculating pH Changes in BuffersThe 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH−(aq) → C2H3O2−(aq) + H2O(l)

HC2H3O2 C2H3O2− OH−

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction 0.280 mol 0.320 mol 0.000 mol

Calculating pH Changes in BuffersNow use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log (0.320)(0. 280)

pH = 4.74 + 0.06

pH = 4.80

PRACTICE EXERCISE

Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl, and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water.

PRACTICE EXERCISE

Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl, and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water.

Answers: (a) 4.68, (b) 1.70


Section 17.3 Titrations

Titration

A known concentration of base (or acid) is slowly added to a solution of acid (or base).

Titration A pH meter or

indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.


Just before and after the equivalence point, the pH increases rapidly.


At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.


As more base is added, the increase in pH again levels off.

SAMPLE EXERCISE 17.6 Calculating pH for a Strong Acid–Strong Base Titration

Calculate the pH when the following quantities of 0.100 M NaOH have been added to 50.0 mL of 0.100 M HCl: (a) 49.0 mL, (b) 51.0 mL.



Solve: The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution (50.0 mL = 0.0500 L) and its molarity (0.100 M):



Solve: The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution (50.0 mL = 0.0500 L) and its molarity (0.100 M):

The number of moles of OH– in 49.0 mL of 0.100 M NaOH is

There are more moles of H+ present than OH–. Each mole of OH– will react with one mole of H+.


The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. At this point in the titration, the solution has a volume of 50.0 mL + 49.0 mL = 99.0 mL. Thus, the concentration of H+(aq) is


The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. At this point in the titration, the solution has a volume of 50.0 mL + 49.0 mL = 99.0 mL. Thus, the concentration of H+(aq) is

The corresponding pH equals

Plan: We are now past the equivalence point and have more OH– in the solution than H+. The initial number of moles of each reactant is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess this time of hydroxide ion.


In this case the total volume of the solution is


Hence, the concentration of OH–(aq) in the solution is



Thus, the pOH of the solution equals



Thus, the pOH of the solution equals

and the pH equals

PRACTICE EXERCISE Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH: (a) 24.9 mL, (b) 25.1 mL.

PRACTICE EXERCISE Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH: (a) 24.9 mL, (b) 25.1 mL.

Answers: (a) 10.30, (b) 3.70

Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.



• The pH at the equivalence point will be >7.



• The pH at the equivalence point will be >7.

• Phenolphthalein is commonly used as an indicator in these titrations.


At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.


With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

The nearly vertical equivalence point portion of the titration curve is smaller for a weak acid-strong base titration, and fewer indicators undergo their color change within this narrow range.

Titration of a Weak Base with a Strong Acid


• The pH at the equivalence point in these titrations is < 7.


• The pH at the equivalence point in these titrations is < 7.

• Methyl red is the indicator of choice.

Titrations of Polyprotic Acids

In these cases there is an equivalence point for each dissociation.

SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 × 10–5).



Plan: We first must determine the number of moles of weak acid and strong base that have been combined. This will tell us how much of the weak acid’s conjugate base has been produced, and we can solve for pH using the equilibrium-constant expression.Solve: Stoichiometry Calculation:



Plan: We first must determine the number of moles of weak acid and strong base that have been combined. This will tell us how much of the weak acid’s conjugate base has been produced, and we can solve for pH using the equilibrium-constant expression.Solve: Stoichiometry Calculation:

The 4.5 × 10–3 mol of NaOH consumes 4.5 × 10–3 mol of HC2H3O2:

The total volume of the solution is


The resulting molarities of HC2H3O2 and C2H3O2

–

after the reaction are therefore



–


Equilibrium Calculation: The equilibrium between HC2H3O2 and C2H3O2

– must obey the equilibrium-constant expression for HC2H3O2:



–




Solving for [H+] gives

We could have solved for pH using the Henderson–Hasselbalch equation.



–




Solving for [H+] gives

PRACTICE EXERCISE(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 × 10–5). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.

PRACTICE EXERCISE(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 × 10–5). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.

Answers: (a) 4.20, (b) 9.26

SAMPLE EXERCISE 17.8 Calculating the pH at the Equivalence Point

Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.


Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.Analyze: Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7.Plan: Determine how many moles of acetic acid there are initially. This will tell us how many moles of acetate ion there will be at the equivalence point. Determine the volume of the solution at the equivalence point and the resultant concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using Kb and the concentration of acetate .



Solve:



Solve:

Hence 5.00 × 10–3 mol of C2H3O2– is formed.

It will take 50.0 mL of NaOH to reach the equivalence point . 50 mL + 50 mL = 100mL = 0.1000 L.



Solve:

Hence 5.00 × 10–3 mol of C2H3O2– is formed.

It will take 50.0 mL of NaOH to reach the equivalence point . 50 mL + 50 mL = 100mL = 0.1000 L.

The C2H3O2– ion is a weak base.

The Kb for C2H3O2– can be calculated from the Ka

value of its conjugate acid, Kb = Kw/Ka = (1.0 × 10–14)/(1.8 × 10–5) = 5.6 × 10–10.





Approximating that and then solving for x, we have x = [OH–] = 5.3 × 10–6 M, which gives pOH = 5.28 and pH = 8.72.Check: The pH is above 7, as expected for the salt of a weak acid and strong base.

PRACTICE EXERCISE

Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (HC7H5O2, Ka = 6.3 × 10–5) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl.

Answers: (a) 8.21, (b) 5.28

PRACTICE EXERCISE

Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (HC7H5O2, Ka = 6.3 × 10–5) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl.


Section 17.4 Solubility EquilibriaSection 17.5 Solubility Factors

• Solubility rules predict that AgCl is insoluble.

• Solubility rules predict that AgCl is insoluble.The term slightly soluble is more accurate

because with very low concentrations of Ag+ and Cl- there is no AgCl(s) formed.



Equilibrium calculations allow us to determine the solubility of these compounds.



Equilibrium calculations allow us to determine the solubility of these compounds.

How much can be in solution before a solid is formedor the maximum solute that can be dissolved in a given amount of solvent.

Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42−]

where the equilibrium constant, Ksp, is called the solubility product.

SAMPLE EXERCISE 17.9 Writing Solubility-Product (Ksp) Expressions

Write the expression for the solubility-product constant for CaF2, and look up the Ksp value in Appendix D.







From Appendix D- this Ksp has a value of 3.9 × 10–11.

PRACTICE EXERCISEGive the solubility-product-constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds: (a) barium carbonate, (b) silver sulfate.

(a) Ksp = [Ba2+][CO32–] = 5.0 × 10–9;

(b) Ksp = [Ag+]2[SO42–] = 1.5 × 10–5

PRACTICE EXERCISEGive the solubility-product-constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds: (a) barium carbonate, (b) silver sulfate.

(a) Ksp = [Ba2+][CO32–] = 5.0 × 10–9;

(b) Ksp = [Ag+]2[SO42–] = 1.5 × 10–5

Solubility Products

Solubility Products• Ksp is not the same as solubility.

Solubility Products• Ksp is not the same as solubility.• Solubility is generally expressed as the

mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

AgCl is most soluble with the highest Ksp

SAMPLE EXERCISE 17.10 Calculating Ksp from Solubility

Solid silver chromate is added to pure water at 25°C. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved.

Analysis of the equilibrated solution shows that its Ag+ concentration is 1.3 × 10–4 M.

Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving the Ag+ or CrO4

2– ions in the solution, calculate Ksp for this compound.

SAMPLE EXERCISE 17.10 Calculating Ksp from SolubilityAnalysis of the equilibrated solution shows that its Ag+ concentration is 1.3 × 10–4 M, calculate Ksp for this compound.

Solve: There are 2 Ag+ ions for each CrO42– ion in solution.

[CrO42–] is half of [Ag+].

[CrO42–] = 1/2 [Ag+]




[CrO42–] = 1/2 [Ag+]




[CrO42–] = 1/2 [Ag+]

Check: We obtain a small value, as expected for a slightly soluble salt.

PRACTICE EXERCISEA saturated solution of Mg(OH)2 in contact with undissolved solid is prepared at 25°C. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions in the solution, calculate Ksp for this compound.

PRACTICE EXERCISEA saturated solution of Mg(OH)2 in contact with undissolved solid is prepared at 25°C. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions in the solution, calculate Ksp for this compound.

Answer: 1.6 × 10–12

SAMPLE EXERCISE 17.11 Calculating Solubility from Ksp

The Ksp for CaF2 is 3.9 × 10–11 at 25°C. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.



The solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is an equilibrium constant.



The solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is an equilibrium constant.

2x moles/liter of F– are produced for each x moles/liter of CaF2 that dissolve.


The Ksp for CaF2 is 3.9 × 10–11 at 25°C. Calculate the solubility of CaF2 in grams per liter.





(Remember that to calculate the cube root of a number, you can use the yx function on your calculator, with )



Solubilities are measured per mole or gram of formula unit (ionic formula). The solubility is “x”.



Solubilities are measured per mole or gram of formula unit (ionic formula). The solubility is “x”.[F-] = 4.2 x 10-4mol/L[Ca2+] = 2.1 x 10-4mol/LThe molar solubility of CaF2 is 2.1 × 10–4 mol/L.



Solubilities are measured per mole or gram of formula unit (ionic formula). The solubility is “x”.[F-] = 4.2 x 10-4mol/L[Ca2+] = 2.1 x 10-4mol/LThe molar solubility of CaF2 is 2.1 × 10–4 mol/L.



[F-] = 4.2 x 10-4mol [Ca2+] = 2.1 x 10-4mol/L The molar solubility of CaF2 is 2.1 × 10–4 mol/L.






Check: We expect a small number for the solubility of a slightly soluble salt.







Check: We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate Ksp: Ksp = (2.1 × 10–4)(4.2 × 10–4)2 = 3.7 × 10–11, close to the Ksp, 3.9 × 10–11 .








Comment: Because F– is the anion of a weak acid, you might expect that the hydrolysis of the ion would affect the solubility of CaF2.








Comment: Because F– is the anion of a weak acid, you might expect that the hydrolysis of the ion would affect the solubility of CaF2. The basicity of F– is so small (Kb = 1.5 × 10–11), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility.








Comment: Because F– is the anion of a weak acid, you might expect that the hydrolysis of the ion would affect the solubility of CaF2. The basicity of F– is so small (Kb = 1.5 × 10–11), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility.The reported solubility is 0.017 g/L at 25°C.




PRACTICE EXERCISEThe Ksp for LaF3 is 2 × 10–19 . What is the solubility of LaF3 in water in moles per liter?

Answer: 9.28 × 10–6 mol/L

PRACTICE EXERCISEThe Ksp for LaF3 is 2 × 10–19 . What is the solubility of LaF3 in water in moles per liter?

Answer: 9.28 × 10–6 mol/L

Factors Affecting Solubility

• The Common-Ion Effect If one of the ions in a solution

equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

BaSO4(s) Ba2+(aq) + SO42−(aq)

SAMPLE EXERCISE 17.12 Calculating the Effect of a Common Ion on Solubility

Calculate the molar solubility of CaF2 at 25°C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF.



In (a) the common ion is Ca2+ and NO3– is a spectator ion.

In (b) the common ion is F– and Na+ is a spectator ion.





Because the slightly soluble compound is CaF2, we need to use the Ksp for this compound, which is available in Appendix D:



The value of Ksp is unchanged by the presence of additional solutes, but the solubility of the salt will decrease in the presence of common ions.



Because the slightly soluble compound is CaF2, we need to use the Ksp for this compound, which is available in Appendix D:

Solve: (a) The initial concentration of Ca2+ is 0.010 M because of the dissolved Ca(NO3)2:



Solve: (a) The initial concentration of Ca2+ is 0.010 M because of the dissolved Ca(NO3)2:



Even without the common-ion effect, the solubility of CaF2 is very small. Assume

Substituting into the solubility-product expression gives







The very small value for x validates the simplifying assumption we have made. Our calculation indicates that 3.1 × 10–5 mol of solid CaF2 dissolves per liter of the 0.010 M Ca(NO3)2 solution.



(b) In this case the common ion is F–, and at equilibrium we have



(b) In this case the common ion is F–, and at equilibrium we have

Assuming that

Thus, 3.9 × 10–7 mol of solid CaF2 should dissolve per liter of 0.010 M NaF solution.



the molar solubility of CaF2 in pure water 2.1 × 10–4 M in 0.010 M Ca2+ 3.1 × 10–5 M in 0.010 M F– 3.9 × 10–7 M.

Much less with common ion!!

More effect because [F-] is squared



PRACTICE EXERCISEThe value for Ksp for manganese(II) hydroxide, Mn(OH)2 , is 1.6 × 10–13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH.

Answer: 4.0 × 10–10 M

PRACTICE EXERCISEThe value for Ksp for manganese(II) hydroxide, Mn(OH)2 , is 1.6 × 10–13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH.


• pH If a substance has a

basic anion, it will be more soluble in an acidic solution.

Substances with acidic cations are more soluble in basic solutions.

SAMPLE EXERCISE 17.13 Predicting the Effect of Acid on Solubility

Which of the following substances will be more soluble in acidic solution than in basic solution: (a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?

Ionic compounds that dissociate to produce a basic anion will be more soluble in acid solution.

(a) Ni(OH)2(s) will be more soluble in acidic solution because of the basicity of OH– ; the H+ ion reacts with the OH– ion, forming water.



(b) Similarly, CaCO3(s) dissolves in acid solutions because CO32– is

a basic anion.

The reaction between CO32– and H+ occurs in a stepwise fashion,

first forming HCO3–. H2CO3 forms in appreciable amounts only

when the concentration of H+ is sufficiently high.



(c) The solubility of BaF2 is also enhanced by lowering the pH, because F– is a basic anion.



(d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.



PRACTICE EXERCISEWrite the net ionic equation for the reaction of the following compounds with acid: (a) CuS, (b) Cu(N3)2.

PRACTICE EXERCISEWrite the net ionic equation for the reaction of the following compounds with acid: (a) CuS, (b) Cu(N3)2.


Factors Affecting Solubility• Complex Ions

Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.



• Complex IonsThe formation of

these complex ions increases the solubility of these salts.

SAMPLE EXERCISE 17.14 Evaluating an Equilibrium Involving a Complex Ion

Calculate the concentration of Ag+ present in solution at equilibrium when concentrated NH3 is added to 0.010M AgNO3 to give an equilibrium concentration of [NH3] = 0.20M. Neglect the small volume change that occurs when NH3 is added.



When NH3(aq) is added to Ag+(aq) a reaction occurs forming Ag(NH3)2

+. Because Kf for the formation of Ag(NH3)2+ is quite large,

we assume that essentially all the Ag+ is converted to Ag(NH3)2+

and approach the problem as though we are concerned with the dissociation of Ag(NH3)2

+ rather than its formation.



When NH3(aq) is added to Ag+(aq) a reaction occurs forming Ag(NH3)2

+. Because Kf for the formation of Ag(NH3)2+ is quite large,

we assume that essentially all the Ag+ is converted to Ag(NH3)2+

and approach the problem as though we are concerned with the dissociation of Ag(NH3)2

+ rather than its formation.

We need to reverse the formation equation to represent the formation of Ag+ and NH3 from Ag(NH3)2

+ and also make thecorresponding change to the equilibrium constant.



If [Ag+] is 0.010 M initially, then [Ag(NH3)2+] will be 0.010 M following

addition of the NH3.




addition of the NH3. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.




addition of the NH3. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.









Solving for x, we obtain x = 1.5 × 10–8 M = [Ag+]. Formation of the Ag(NH3)2

+ complex drastically reduces the concentration of free Ag+ ion in solution.



PRACTICE EXERCISECalculate [Cr3+] in equilibrium with Cr(OH)4

– when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0.

PRACTICE EXERCISECalculate [Cr3+] in equilibrium with Cr(OH)4

– when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0.

Answer: 1 × 10–16 M


• Amphoterism Amphoteric metal

oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.


• Amphoterism Amphoteric metal

oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

Examples of such cations are Al3+, Zn2+, and Sn2+.

They are insoluble in water but dissolve readily in the presence of an acid or base.

They are insoluble in water but dissolve readily in the presence of an acid or base.


Section 17.6 & 17.7Selective Precipitation/ Qualitative Analysis

Will a Precipitate Form?

• In a solution,If Q = Ksp, the system is at equilibrium and the

solution is saturated.



solution is saturated.If Q < Ksp, more solid will dissolve until Q = Ksp.



solution is saturated.If Q < Ksp, more solid will dissolve until Q = Ksp.If Q > Ksp, the salt will precipitate until Q = Ksp.

SAMPLE EXERCISE 17.15 Predicting Whether a Precipitate Will Form

Will a precipitate form when 0.10 L of 8.0 × 10–3 M Pb(NO3)2 is added to 0.40 L of 5.0 × 10–3 M Na2SO4?



Plan: Determine the concentrations of all ions immediately upon mixing of the solutions and compare the value of the reaction quotient, Q, to the solubility-product constant, Ksp, for any potentially insoluble product.




The possible metathesis products are PbSO4 and NaNO3.





Sodium salts are quite soluble;





Sodium salts are quite soluble;

PbSO4 has a Ksp of 6.3 × 10–7 (Appendix D), however, and will precipitate if the Pb2+ and SO4

2– ion concentrations are high enough for Q to exceed Ksp for the salt.

When the two solutions are mixed, the total volume is 0.10 L + 0.40 L = 0.50 L.




The number of moles of Pb2+ is




The number of moles of Pb2+ is

The concentration of Pb2+ in the 0.50-L mixture is therefore



The number of moles of SO42– is




The number of moles of SO42– is

Therefore, [SO42–] in the 0.50-L mixture is








Because Q > Ksp, PbSO4 will precipitate.



PRACTICE EXERCISEWill a precipitate form when 0.050 L of 2.0 × 10–2 M NaF is mixed with 0.010 L of 1.0 × 10–2 M Ca(NO3)2?

PRACTICE EXERCISEWill a precipitate form when 0.050 L of 2.0 × 10–2 M NaF is mixed with 0.010 L of 1.0 × 10–2 M Ca(NO3)2?

yes, CaF2 precipitates because Q = 4.6 × 10–8 is larger than Ksp = 3.9 × 10–11

SAMPLE EXERCISE 17.16 Calculating Ion Concentrations for Precipitation

A solution contains 1.0 × 10–2 M Ag+ and 2.0 × 10–2M Pb2+. When Cl– is added to the solution, both AgCl (Ksp = 1.8 × 10–10) and PbCl2 (Ksp = 1.7 × 10–5) precipitate from the solution. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?



Plan: We can calculate what concentration of Cl– ion would be necessary to begin precipitation of each. The salt requiring the lower Cl– ion concentration will precipitate first.

[Ag+] = 1.0 × 10–2 M



[Ag+] = 1.0 × 10–2 M

Any Cl– in excess of 1.8 x 10-10 will cause AgCl to precipitate.









[Cl–] in excess of 2.9 × 10–2 M will cause PbCl2 to precipitate.

[Cl–] in excess of 1.8 x 10-10 will cause AgCl to precipitate.[Cl–] in excess of 2.9 × 10–2 M will cause PbCl2 to precipitate.




AgCl will precipitate first because it requires a much smaller concentration of Cl–.




AgCl will precipitate first because it requires a much smaller concentration of Cl–.

Ag+ can be separated from Pb2+ by slowly adding Cl– so [Cl–] is between 1.8 × 10–8 M and 2.9 × 10–2 M.



PRACTICE EXERCISEA solution consists of 0.050 M Mg2+ and 0.020 M Cu2+.

Which ion will precipitate first as OH– is added to the solution?

What concentration of OH– is necessary to begin the precipitation of each cation? Ksp = 1.8 × 10–11 for Mg(OH)2Ksp = 2.2 × 10–20 for Cu(OH)2.

PRACTICE EXERCISEA solution consists of 0.050 M Mg2+ and 0.020 M Cu2+.

Which ion will precipitate first as OH– is added to the solution?

What concentration of OH– is necessary to begin the precipitation of each cation? Ksp = 1.8 × 10–11 for Mg(OH)2Ksp = 2.2 × 10–20 for Cu(OH)2.

Answer: Cu(OH)2 precipitates first. Cu(OH)2 begins to precipitate when [OH–] exceeds 1.0 × 10–9 M; Mg(OH)2 begins to precipitate when [OH–] exceeds 1.9 × 10–5 M.

Cu2+(aq) + H2S(aq) ⇌ CuS(s) + 2 H+(aq)

A high concentration of H2S and a low concentration of H+ (high pH) will reduce [Cu2+].

Cu2+(aq) + H2S(aq) ⇌ CuS(s) + 2 H+(aq)

Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

The solution must contain one of the following cations:

Ag+, Pb2+ or Hg22+.

Education

Chapter 17 Lecture- Buffers & Ksp