15 Burdge Lecture Notes.pdf

(1)

Chemistry: Atoms First

Julia Burdge & Jason Overby

Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 15

Acid-Base Equilibria and

Solubility Equilibria

Kent L. McCorkle Cosumnes River College

Sacramento, CA

Acid-Base Equilibria and Solubility Equilibria

17

17.1 The Common Ion Effect 17.2 Buffer Solutions

Calculating the pH of a Buffer

Preparing a Buffer Solution with as Specific pH 17.3 Acid-Base Titrations

Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Strong Acid-Weak base Titrations Acid-Base Indicators 17.4 Solubility Equilibria

Solubility Product Expression and Ksp Calculations Involving Ksp and Solubility Predicting Precipitation Reactions 17.5 Factors Affecting Solubility

The Common Ion Effect pH

Complex Ion Formation

17.6 Separation of Ions Using Differences in Solubility Fractional Precipitation

Qualitative Analysis of Metal Ions in Solution

The Common Ion Effect

A system at equilibrium will shift in response to being stressed.

The addition of a reactant or a product can be an applied stress.

[H

+

] = [CH

₃

COO

–

] = 1.34 x 10

–3

M; pH = 2.87

17.1

Initial concentration (M) 0.10 0 0

Change in concentration (M) –x +x +x

Equilibrium concentration (M) 0.10 – x x x

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

CH3COONa(aq) H2O Na+₍aq) + CH3COO–(aq)

CH3COOH(aq) H+₍aq) + CH3COO–(aq)

Equilibrium is driven toward reactant

addition

Worked Example 17.1

Strategy Construct a new equilibrium table to solve for the hydrogen ion concentration. We use the stated concentration of acetic acid, 0.10 M, and [H+_{] ≈ 0}_M_{as the initial concentrations in the table.}

Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10-M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.)

Initial concentration (M) 0.10 0 0.050

Change in concentration (M) –x +x +x Equilibrium concentration (M) 0.10 – x x 0.050 + x

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

Worked Example 17.1 (cont.)

Solution These equilibrium concentrations are then substituted into the equilibrium expression to give

Because we expect x to be very small (even smaller than 1.34×10-3_M_–see

above), because the ionization of CH3COOH is suppressed by the presence of

CH3COO-, we assume

(0.10 – x) M ≈ 0.10 M and (0.050 + x) M ≈ 0.050 M

Therefore, the equilibrium expression simplifies to

and x = 3.6×10-5_M_{. According to the equilibrium table, [H}+_{] =}_x_{, so}

pH = –log(3.6×10-5_{) = 4.44.}

1.8×10-5₌(x)(0.050 + x)

0.010 – x

1.8×10-5₌(x)(0.050)

0.010

Worked Example 17.1 (cont.)

Think About It The equilibrium concentrations of CH3COOH, CH3COO-, and

H+_{are the same regardless of whether we add sodium acetate to a solution of}

acetic acid, add acetic acid to a solution of sodium acetate, or dissolve both species at the same time. We could have constructed an equilibrium table starting with the equilibrium concentrations in the 0.10 M acetic acid solution:

In this case, the reaction proceeds to the left. (The acetic acid concentration increases, and the concentrations of hydrogen and acetate ions decrease.) Solving for y gives 1.304×10-3_M_{. [H}+_{] = 1.34}×10-3_{– y}_{= 3.6}×10-5_M_{and pH = 4.44.}

We get the same pH either way.

Initial concentration (M) 0.09866 1.34×10-3 _5.134×10-2

Change in concentration (M) +y –y –y Equilibrium concentration (M) 0.09866 + y 1.34×10-3_–_{y 5.134}×10-2 –

(2)

Buffer Solutions

17.2

A solution that contains a weak acid and its conjugate base (or a

weak base and its conjugate acid) is a buffer.

Buffer solutions resist changes in pH.

CH

3

COOH(

aq

)

⇌

H

+

(

aq

) + CH

3

COO

–

(

aq

)

acid conjugate base

reacts with

added base reacts with added acid

CH

3

COOH(

aq

) +

OH

–

(

aq

)

CH

3

COO

–

(

aq

) + H

2

O(

l

)

CH

3

COOH(

aq

)

CH

3

COO

–

(

aq

) +

H

+

(

aq

)

Buffer Solutions

Calculate the pH of a buffer that contains 1.0 M acetic acid and 1.0

M sodium acetate.

x = [H

+

] = 1.8 x 10

–5

M; pH = 4.74

Change in concentration (M)

Equilibrium concentration (M)

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

–x +x +x

1.0 – x x 1.0 + x

 



5

a

1.0

1.8 10

x

K

x











Buffer Solutions

Calculate the pH of the same buffer (1.0 M acetic acid and 1.0 M

sodium acetate) after the addition of 0.10 mol of HCl.

The added acid reacts with the conjugate base (acetate ion).

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

–x +x +x

1.1 – x x 0.9 + x

CH3COO–(aq) + H+(aq) CH3COOH(aq)

Upon addition of H+_: _{1.0 mol} _{0.1 mol} _{1.0 mol}

After H+_{has been consumed: 0.9 mol} _{0 mol} _{1.1 mol}

Buffer Solutions

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

–x +x +x

1.1 – x x 0.9 + x

 



5

a

1.1

0.9

1.8 10

x

K

x











x = [H

+

] = 3.6 x 10

–5

M;

pH = 4.66

The original buffer had a pH = 4.74

Buffer Solutions

The pH of a buffer solution can often be calculated with the

Henderson-Hasselbalch equation.









a

conjugate base

pH p log

weak acid

K





Worked Example 17.2

Strategy Added base will react with the acetic acid component of the buffer, converting OH-_{to CH}₃_COO-_:

CH3COOH(aq) + OH-(aq) ⇌ H2O(l) + CH3COO–(aq)

Write the starting amount of each species above the equation and the final amount of each species below the equation. Use the final amounts as concentrations in pH = pKa + log([A-]/[HA]).

Upon addition of OH-_: _{1.00 mol} _{0.10 mol} _{1.00 mol}

After OH-_consumed: _{0.90 mol} _{0 mol} _{1.10 mol}

(3)

Worked Example 17.2 (cont.)

Solution These equilibrium concentrations are then substituted into the equilibrium expression to give

Thus, the pH of the buffer after addition of 0.10 mole of NaOH is 4.83. pH = 4.74 + log1.10 _0.90M_M

pH = 4.83

Think About It Always do a “reality check” on a calculated pH. Although a buffer does minimize the effect of added base, the pH does increase. If you find that you’ve calculated a lower pH after the addition of a base, check for errors like mixing up the weak acid and conjugate base concentrations or losing track of a minus sign.

Buffer Solutions

Buffers must have conjugate acid and base concentrations within a

factor of 10.

The pH of a buffer cannot be more than one pH unit different than

the pK

a

of the weak acid it contains.









conjugate base

weak acid

10



0.1

Buffer Solutions

To make a buffer with a specific pH:

1) Pick a weak acid whose pK

a

is close to the desired pH.

2) Substitute the pH and pK

a

into the equation below to obtain the

necessary [conjugate base]/[weak acid] ratio.

 

a

A pH = pK log_HA



 

 



Worked Example 17.3

Strategy Select an acid with a pKa within one pH unit

of 9.50. Use the pKa of the acid and pH = pKa + log([A

-]/[HA]) to calculate the necessary ratio of [conjugate base]/[weak acid]. Select concentrations of the buffer components that yield the calculated ratio. Select an appropriate weak acid from the table at right, and describe how you would prepare a buffer with a pH of 9.50.

Weak Acid Ka pKa

HF 7.1×10-4 _3.15 HNO2 4.5×10-4 3.35 HCOOH 1.7×10-4 _3.77 C6H5COOH 6.5×10-5 _4.19 CH3COOH 1.8×10-5 4.74 HCN 4.9×10

-10 9.31 C6H5OH 1.3×10

-10 9.89

Solution Two of the acids listed have pKa values in the desired range:

hydrocyanic acid (HCN, pKa = 9.31) and phenol (C6H5OH, pKa = 9.89).

9.50 = 9.89 + log[C6H5O-]

[C6H5OH]

9.50 – 9.89 = log = –0.39[C6H5O-]

[C6H5OH]

[C6H5O-]

[C6H5OH] = 10 -0.39_{= 0.41}

Worked Example 17.3 (cont.)

Solution Therefore, the ratio of [C6H5O-] to [C6H5OH] must be 0.41 to 1. One

way to achieve this would be to dissolve 0.41 mole of C6H5ONa and 1.00 mole of

C6H5OH in 1.00 L of water.

Think About It There is an infinite number of combinations of [conjugate base] and [weak acid] that will give the necessary ratio. Note that this pH could also be achieved using HCN and a cyanide salt. For most purposes, it is best to use the least toxic compounds available.

Acid-Base Titrations

Strong Acid-Strong Base Titrations

The reaction between the strong acid HCl and the strong base NaOH

can be represented by:

NaOH(aq) + HCl(aq) → NaCl(aq) + H

2

O(l)

or by the net ionic equation,

(4)

Acid-Base Titrations

Titration of 25.0 mL of 0.100 M HCl with

0.100 M NaOH

Acid-Base Titrations

Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH

Weak Acid-Strong Base Titrations:

Consider the neutralization between acetic acid and sodium

hydroxide:

CH

3

COO

–

(aq) + H

2

O(l)(aq)

⇌

CH

3

COOH(aq) + OH

–

(aq)

CH

3

COOH(aq) + OH

–

(aq) → CH

3

COO

–

(aq) + H

2

O(l)

The acetate ion that results from this neutralization undergoes

hydrolysis:

Weak Acid-Strong Base Titrations

Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH

The initial pH is determined by the ionization of acetic acid.

x = [H

+

] = 1.34 x 10

–3

M; pH = 2.87

CH3COOH(aq) ⇌ H+(aq) + CH 3COO–(aq)

–x +x +x

0.10 – x x x

2

5 a

0.10

1.8 10

x

K

x









Titration of 25.0 mL of 0.100 M acetic

acid with 0.100 M NaOH

Acid-Base Titrations

After the addition of base, some of the acetic acid has been

converted to acetate ion:

The solution is a buffer and the Henderson-Hasselbalch equation

can be used to calculate pH.

After 10.0 mL of base has been added:

CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)

 

a

A pH = p log

HA

K



 

 

 pH = 4.74 log



1.0 mmol



4.56 1.5 mmol

 

Volume of OH–

added (mL) OH

–_added

(mol) CHremaining 3COOH CH3COO

–

produced pH

10.0 1.0 1.5 1.0 4.56

(5)

Acid-Base Titrations

At the equivalence point, all the acetic acid has been neutralized.

CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)

Volume of OH–

added (mL) OH

–_added

–

produced pH

25.0 2.5 0.0 2.5 8.72

Change in concentration (M) –x +x +x Equilibrium concentration (M) 0.050 – x x x

CH3COO–(aq) + H2O(l) ⇌ OH–₍aq) + CH3COOH(aq)

Must use TOTAL VOLUME to calculate concentration. 3

2.5 mmol CH COO 0.050

50.0 mL M 

   

 

Use K

a

x K

b

= K

w

to get K

b

for the acetate ion; K

b

= 5.6 x 10

–10

x = [OH

–

] = 5.3 x 10

–6

; pOH = 5.28; pH = 8.72

CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH(aq)





2

3 10

b 3 OH CH COOH

5.6 10 0.050 CH COO

x

K _x



 

   

   



 

 

Acid-Base Titrations

After the equivalence point, all the acetic acid has been neutralized, nothing is left to neutralize the added strong base.

Volume of

OH–_added

(mL)

OH–

added (mmol)

Excess OH– (mmol)

Total Volume

(mL) [OH–_]

(mol/L) pOH pH

30.0 3.0 0.5 55.0 0.0091 2.04 11.96 35.0 3.5 1.0 60.0 0.017 1.78 12.22

Volume of OH–

added (mL) OH

–_added

–

produced pH

25.0 2.5 0.0 2.5 8.72

Titration of 25.0 mL of 0.100 M acetic

acid with 0.100 M NaOH

Acid-Base Titrations

Weak Acid-Strong Base Titrations

Worked Example 17.4

Strategy The reaction between acetic acid and sodium hydroxide is

OH–₍_aq_{) + CH}₃_COOH₍_aq₎⇌ CH3COO–(aq) + H2O(l)

Prior to the equivalence point [part (a)], the solution contains both acetic acid and acetate ion, making the solution a buffer. We can solve part (a) using the Henderson-Hasselbach equation. At the equivalence point [part (b)], all the acetic acid has been neutralized and we have only acetate ion in solution. We must determine the concentration of acetate ion and solve part (b) as an equilibrium problem, using the Kb for acetate ion. After the equivalence point [part (c)], all the

acetic acid has been neutralized and there is nothing to consume the additional added base. We must determine the concentration of excess hydroxide ion in the solution and solve for pOH = -log[OH-_{] and pH + pOH = 14.00.}

Calculate the pH in the titration of 50.0 mL of 0.120 M acetic acid by 0.240 M

sodium hydroxide after the addition of (a) 10.0 mL of base, (b) 25.0 mL of base, and (c) 35.0 mL of base.

Worked Example 17.4 (cont.)

Solution Remember that M can be defined as either mol/L or mmol/mL. For this type of problem, it simplifies the calculations to use millimoles rather than moles. Ka for acetic acid is 1.8×10-5, so pKa = 4.74. Kb for acetate ion is

5.6×10-10_.

(a) The solution originally contains (0.120 mmol/mL)(50.0 mL) = 6.00 mmol of acetic acid. A 10.0-mL amount of base contains (0.240 mmol/mL)(10.0 mL) = 2.40 mmol of base. After the addition of 10.0 mL of base, 2.40 mmol of OH-_has

neutralized 2.40 mmol of acetic acid, leaving 3.60 mmol of acetic acid and 2.40 mmol acetate ion in solution.

Upon addition of OH-_: _{6.00 mol} _{2.40 mol} _{0 mol}

After OH-_consumed: _{3.60 mol} _{0 mol} _{2.40 mol}

(6)

Worked Example 17.4 (cont.)

Solution (b) After the addition of 25.0 mL of base, the titration is at the equivalent point. We calculate the pH using the concentration and the Kb of

acetate ion.

At the equivalence point, we have 6.0 mmol of acetate ion in the total volume. We determine the total volume by calculating what volume of 0.24 M contains 6.0 mmol:

(volume)(0.240 mmol/mL) = 6.00 mmol

volume = = 25.0 mL

Therefore, the equivalence point occurs when 25.0 mL of base has been added, making the total volume 50.0 mL + 25.0 mL = 75.0 mL. The concentration of acetate ion at the equivalence point is therefore

= 0.0800 M

6.00 mmol 0.240 mmol/mL

6.00 mmol CH3COO-

75.0 mL

Worked Example 17.4 (cont.)

Solution (b) We can construct an equilibrium table using this concentration and solve for pH using the ionization constant for CH3COO- (Kb = 5.6×10-10):

Using the equilibrium expression and assuming that x is small enough to be neglected,

According to the equilibrium table, x = [OH-_{], so [OH}-_{] = 6.7}×10-6_M_{. At}

equilibrium pOH = – log(6.7×10-6_{) = 5.17 and pH = 14.00 – 5.17 = 8.83.}

CH3COO-(aq) + H2O(l) ⇌ OH-(aq) + CH3COOH(aq)

Kb =[CH3COOH][OH -_]

[CH3COO-] = 1.8×10

-5

=_{0.0800 –}(x)(x)_x

x = = 6.7₄_.₄₈_₁₀11 ×10-6_M

≈_0.0800x2

Acid-Base Titrations

Strong Acid-Weak Base Titrations

Titration of 25.0 mL of 0.100 M NH3 with

0.100 M HCl

Worked Example 17.5

Strategy The reaction between NH3 and HCl is

NH3(aq) + H+(aq) ⇌ NH4+(aq)

At the equivalence point, all the NH3 has been converted to NH4+. Therefore, we

must determine the concentration of NH4+ at the equivalence point and use the Ka

for NH4+ to solve for pH using an equilibrium table.

Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is

titrated with 0.100 M HCl.

Solution The solution originally contains (0.100 mmol/mL)(25.0 mL) = 2.50 mmol NH4+. At the equivalence point, 2.50 mmol of HCl has been added. The

volume of 0.100 M HCl that contains 2.50 mmol is

(volume)(0.100 mmol/mL) = 2.50 mmol

volume = = 25.0 mL _{0.100 mmol/mL}2.50 mmol

Worked Example 17.5 (cont.)

Solution It takes 25.0 mL of titrant to reach the equivalence point, so the total solution is 25.0 + 25.0 = 50.0 mL. At the equivalence point, all the NH3 originally

present has been converted to NH4+. The concentration of NH4+ is

(2.50 mmol)/(50.0 mL) = 0.0500 M. We must use this concentration as the starting concentration of ammonium ion in our equilibrium table.

The equilibrium expression is

NH4+₍_aq_{) + H2O(}_l₎⇌ NH3(aq) + H3O+₍_aq₎

Ka =[NH3][H +_]

[NH4+] = 5.6×10

-10

=_{0.0500 –}(x)(x)_x

x = = 5.3₂_.₈_₁₀11 ×10-6_M_{= [H}+_]

≈_0.0500x2

pH = –log(5.3×10-6_{) = 5.28} Think About It In the titration of a weak base with a strong acid, the species in solution at the equivalence point is the conjugate acid. Therefore, we should expect an acidic pH. Once all the NH3 has

been converted to NH4+, there is no longer anything in the solution

to consume the added acid. Thus, the pH after the equivalence point depends on the number of millimoles of H+_{added and not consumed}

divided by the new total volume.

Acid-Base Titrations

The equivalence point in a titration can be visualized with the use

of an acid-base indicator.

The endpoint of a titration is the

point at which the color of the

indicator changes.

(7)

Worked Example 17.6

Strategy Determine the pH range that corresponds to the steepest part of each titration curve and select an indicator (or indicators) that changes color within that range.

Which indicator listed in Table 17.3 would you use for the acid-base titrations shown in (a) Figure 17.3, (b) Figure 17.4, and (c) Figure 17.5?

Solution (a) The titration curve at right is for the titration of a strong acid with a strong base. The steep part of the curve spans a pH range of about 4 to 10.

Most of the indicators in Table 17.3, with the exceptions of thymol blue, bromophenol blue, and methyl orange, would work for the titration of a strong acid with a strong base.

Worked Example 17.6 (cont.)

Solution (b) The figure at right shows the titration of a weak acid with a strong base. The steep part of the curve spans a pH range of about 7 to 10.

Cresol red and phenolphthalein are suitable indicators.

(c) The figure at right shows the titration of a weak base with a strong acid. The steep part of the curve spans a pH range of 7 to 3.

Bromophenol blue, methyl orange, methyl red, and chlorophenol blue are all suitable indicators.

Think About It If we don’t select an appropriate indicator, the endpoint (color change) will not coincide with the equivalence point.

Solubility Equilibria

Quantitative predictions about how much of a given ionic compound

will dissolve in water is possible with the solubility product

constant, K

sp.

17.4

AgCl(

s

)

⇌

Ag

+

(

aq

) + Cl

–

(

aq

)

K

sp

= [Ag

+

][Cl

–

]

Compound Dissolution Equilibrium Ksp Aluminum hydroxide Al(OH)3(s) ⇌ Al3+(aq) + 3OH–(aq) 1.8 x 10–33

Calcium fluoride CaF2(s) ⇌ Ca2+(aq) + 2F–(aq) 4.0 x 10–11

Silver bromide AgBr(s) ⇌ Ag+_{(aq) + Br}–_(aq) _{7.7 x 10}–13

Silver chloride AgCl(s) ⇌ Ag+_{(aq) + Cl}–_(aq) _{1.6 x 10}–6

Zinc sulfide ZnS(s) ⇌ Zn2+_{(aq) + S}2–_(aq) _{3.0 x 10}–23

Solubility Equilibria

Molar solubility

is the number of moles of solute in 1 L of a

saturated solution (mol/L)

Solubility is the number of grams of solute in 1 L of a saturated

solution (g/L).

To calculate a compound’s molar solubility:

1)

Construct an equilibrium table.

2)

Fill in what is known.

(8)

Solubility Equilibria

The K

sp

of silver bromide is 7.7 x 10

–13

. Calculate the molar

solubility.

Initial concentration (M) 0 0

Change in concentration (M) +s +s

Equilibrium concentration (M) s s

AgBr(s) ⇌ Ag+(aq) + Br–(aq)

K

sp

= [Ag

+

][Br

–

] = 7.7 x 10

–13

7.7 x 10

–13

= (s)(s)

s = 8.8 x 10

–7

M

7

4

8.8 10 mol AgBr 187.8 g _{1.7 10 g/L} 1 L 1 mol AgBr



 _ _ _

Worked Example 17.7

Strategy Write the dissociation equation for Cu(OH)2, and look up its Ksp value

in Table 17.4. Solve for molar solubility using the equilibrium expression. Convert molar solubility to solubility in g/L using the molar mass of Cu(OH)2.

Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in g/L.

Solution The equation for the dissociation of Cu(OH)2 is

Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq)

and the equilibrium expression is Ksp = [Cu2+][OH-]2. According to Table 17.4, Ksp for Cu(OH)2 is 2.2×10-20. The molar mass of Cu(OH)2 is 97.57 g/mol.

Initial concentration (M) 0 0

Change in concentration (M) +s +2s

Equilibrium concentration (M) s 2s

Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq)

Worked Example 17.7 (cont.)

Solution Therefore,

2.2×10-20_{= (}_s₎₍₂_s₎2_{= 4}_s3

s = = 1.8×10-7_M

The molar solubility of Cu(OH)2 is 1.8×10-7M. Multiplying by its molar mass

gives

solubility of Cu(OH)2 = 1.8×10 -7

mol L = 1.7×10-5_g/L

3 20

4 10 2 . 2 _ 

×97.57 g Cu(OH)_{1 mol Cu(OH)}2 2

Think About It Common errors arise in this type of problem when students neglect to raise an entire term to the appropriate power. For example, (2s)2_is

equal to 4s2_{(not 2}_s2_).

Worked Example 17.8

Strategy Convert solubility to molar solubility using the molar mass of CaSO4,

and substitute the molar solubility into the equilibrium expression to determine Ksp.

The solubility of calcium sulfate (CaSO4) is measured experimentally and found

to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate.

Solution The molar mass of CaSO4 is 136.2 g/mol. The molar solubility of

CaSO4 is

The equation and the equilibrium expression for the dissociation of CaSO4 are

CaSO4(s) ⇌ Ca2+(aq) + SO42-(aq) and Ksp = [Ca2+][SO42-]

substituting the molar solubility into the equilibrium expression gives

Ksp = (s)(s) = (4.9×10-3)2 = 2.4×10-5

molar solubility of CaSO4 = 0.67 g CaSO_{1 L} 4 s = 4.9×10-3_mol/L

× 1 mol CaSO4

136.2 g CaSO4 Think About It The Ksp for CaSO4 is relatively large (compared to

many of the Ksp values in Table 17.4). In fact, sulfates are listed as

soluble compounds in Table 9.2, but calcium sulfate is listed as an insoluble exception. Remember that the term insoluble really refers to compounds that are slightly soluble, and that different sources may differ with regard to how soluble a compound must be to be considered soluble.

Solubility Equilibria

For the dissociation of an ionic solid in water, the following

conditions may exist:

1)

The solution is unsaturated

2)

The solution is saturated

3)

The solution is supersaturated

The following relationships are useful in making predictions on

when a precipitate might form.

(9)

Worked Example 17.9

Strategy For each part, identify the compound that might precipitate and look up its Ksp value in Table 17.4 or Appendix 4. Determine the concentration of each

compound’s constituent ions, and use them to determine the value of the reaction quotient, Qsp; then compare each reaction quotient with the value of the

corresponding Ksp. If the reaction quotient is greater than Ksp, a precipitate will

form.

Predict whether a precipitate will form when each of the following is added to 650 mL of 0.080 M K2SO4: (a) 250 mL of 0.0040 M BaCl2; (b) 175 mL of 0.15 M

AgNO3; (c) 325 mL of 0.25 M Sr(NO3)2. (Assume volumes are additive.)

Solution (a) BaSO4 might form and its Ksp = 1.1×10-10. Concentrations of the

constituent ions of BaSO4 are:

Using these concentrations in the equilibrium expression, [Ba2+_][SO 42-], gives a

reaction quotient of (0.0011)(0.0058) = 6.4×10-6_{, which is greater than the} Ksp of BaSO4 (1.1×10-10). Therefore, BaSO4 will precipitate.

[Ba2+_{] = 250 mL}× 0.0040 M 650 mL + 250 mL

= 0.0011 M [SO42-_{] = 650 mL}× 0.0080 M 650 mL + 250 mL

= 0.0058 M

Worked Example 17.9 (cont.)

Solution (b) Ag2SO4 might form and its Ksp = 1.5×10-5. Concentrations of the

constituent ions of Ag2SO4 are:

Using these concentrations in the equilibrium expression, [Ag+_]2_[SO₄2-_{], gives a}

reaction quotient of (0.032)2_{(0.0063) = 6.5}×10-6_{, which is less than the}_K_sp_of

Ag2SO4 (1.5×10-5). Therefore, Ag2SO4 will not precipitate.

(c) SrSO4 might form and its Ksp = 3.8×10-7. Concentrations of the constituent

ions of Ag2SO4 are:

Using these concentrations in the equilibrium expression, [Sr2+_][SO 42-], gives a

reaction quotient of (0.083)(0.0053) = 4.4×10-4_{, which is greater than the}_K_sp_of

SrSO4 (3.8×10-7). Therefore, SrSO4 will precipitate.

[Ag+] = 175 mL _{650 mL + 175 mL = 0.0032}× 0.15 M M [SO42-] = 650 mL _M× 0.0080 650 mL + 175 mL

= 0.0063 M

[Sr2+_{] = 325 mL}× 0.25 M

650 mL + 325 mL = 0.083 M [SO42-] = 650 mL _M× 0.0080 650 mL + 325 mL

= 0.0053 M Think About It Students sometimes have difficulty deciding what compound might precipitate. Begin by writing down the constituent ions in the two solutions before they are combined. Consider the two possible combinations; the cation from the first solution and the anion from the second, or vice versa. You can consult the information in Table 9.2 and 9.3 to determine whether one of the combinations is insoluble. Also keep in mind that only an insoluble

salt will have a tabulated Ksp value.

Factors Affecting Solubility

Several factors exist that affect the solubility of ionic compounds:



Common ion effect



pH



Formation of complex ions

17.5

Worked Example 17.10

Strategy Silver nitrate is a strong electrolyte that dissociates completely in water. Therefore, the concentration of Ag+_{before any AgCl dissolves is 6.5}×10-3 M. Use the equilibrium expression, the Ksp for AgCl, and an equilibrium table to

determine how much AgCl will dissolve.

Calculate the molar solubility of silver chloride in a solution that is 6.5×10-3_M

in silver nitrate.

Solution The dissolution equilibrium and the equilibrium expression are AgCl(s) ⇌ Ag+₍_aq_{) + Cl}-₍_aq₎

Initial concentration (M) 6.5×10-3 ₀

Change in concentration (M) +s +s

Equilibrium concentration (M) 6.5×_s10-3 + s

AgCl(s) ⇌ Ag+₍_aq_{) + Cl}-₍_aq₎ Ksp = [Ag+][Cl-]

Worked Example 17.10 (cont.)

Solution Substituting these concentrations into the equilibrium expression gives 1.6×10-10_{= (6.5}×10-3₊_s₎₍_s₎

We expect s to be very small, so

6.5×10-3₊_{s ≈}_6.5×10-3

and

1.6×10-10_{= (6.5}×10-3₎₍_s₎

Thus

Therefore, the molar solubility of AgCl in 6.5×10-3_M_AgNO₃_{is 2.5}×10-8_M_. s = 1.6×1010

-6.5×10-3

= 2.5×10-8_M

Think About It The molar solubility of AgCl in water is = 1.3×10-5 M. The presence of 6.5×10-3_M_{AgNO3 reduces the solubility of AgCl by a}

factor of ~500.

10

10 6 . 1 _ 

Factors Affecting Solubility

pH:

K

sp

= [Mg

2+

][OH

–

]

2

= 1.2 x 10

–11

(s)(2s)

2

= 4s

3

= 1.2 x 10

–11

s = 1.4 x 10

–4

M

At equilibrium:

[OH

–

] = 2(1.4 x 10

–4

M ) = 2.8 x 10

–4

M

pOH = –log(2.8 x 10

–4

) = 3.55

pH = 14.00 – 3.55 = 10.45

In a solution with a pH of less than 10.45, the solubility of Mg(OH)

2

increases.

(10)

Factors Affecting Solubility

pH:

If the pH of the medium were higher than 10.45, [OH

–

] would be

higher and the solubility of Mg(OH)

2

would decrease because of the

common ion (OH

-

) effect.

Mg(OH)

2

(s)

⇌

Mg

2+

(aq) + 2OH

–

(aq)

Mg(OH)

2

(s) + 2H

+

(aq)

⇌

Mg

2+

(aq) + 2H

2

O(l)

2H

+

(aq) + 2OH

–

(aq) → 2H

2

O(l)

Overall:

Factors Affecting Solubility

pH:

As the concentration of F

–

decreases, the concentration of Ba

2+

must

increase so satisfy the equality:

K

sp

= [Ba

2+

][F

–

]

2

The solubilities of salts containing anions that do not hydrolyze are

unaffected by pH:

Cl

–

, Br

–

, NO

₃–

BaF

2

(s)

⇌

Ba

2+

(aq) + 2F

–

(aq)

BaF

2

(s) + 2H

+

(aq)

⇌

Ba

2+

(aq) + 2HF(aq)

2H

+

(aq) + 2F

–

(aq) → 2HF(aq)

Overall:

Strategy For each salt, write the dissociation equilibrium equation and determine whether it produces an anion that will react with H+_{. Only an anion that}

is the conjugate base of a weak acid will react with H+_.

Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO4?

Solution The dissolution equilibrium and the equilibrium expression are (a) CuS(s) ⇌ Cu2+₍_aq_{) + S}2-₍_aq₎

S2-_{is the conjugate base of the weak acid HS}-_{. S}2-_{reacts with H}+_{as follows:}

S2-₍_aq_{) + H}+₍_aq₎⇌ HS-₍_aq₎

(b) AgCl(s) ⇌ Ag+₍_aq_{) + Cl}-₍_aq₎

Cl-_{is the conjugate base of the strong acid HCl. Cl}-_{does not react with H}+_.

(c) PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)

SO42- is the conjugate base of the weak acid HSO4-. It reacts with H+ as follows:

SO42-(aq) + H+(aq) ⇌ HSO4-(aq)

Worked Example 17.11 (cont.)

Solution CuS and PbSO4 are more soluble in acid than in water. (AgCl is no

more or less soluble than in water.)

Think About It When a salt dissociates to give the conjugate base of a weak acid, H+_{ions in an acidic solution}_consume_{a product (base) of the dissolution.}

This drives the equilibrium to the right (more solid dissolves) according to Le Châtelier’s principle.

Factors Affecting Solubility

Complex ion formation:

A complex ion

is an ion containing a central

metal cation bonded to one or more

molecules or ions.

A solution of CoCl

2

is pink because of the

presence of Co(H

2

O)

62+

ions.

When HCl is added to a CoCl

2

solution, there is

a color change from pink to blue.

(11)

Cu(OH)

2

(s) + 4NH

3

(aq)

⇌

Cu(NH

3

)

42+

(aq) + 2OH

–

(aq)

The formation of the Cu(NH

3

)

42+

ion can be expressed as

Cu

2+

(aq) + 2OH

–

(aq) → Cu(OH)

2

(s)

Cu

2+

(aq) + 4NH

₃

(aq)

⇌

Cu(NH

₃

)

₄2+

(aq)









2

3 4 13

f 2 4

3 Cu NH

5.0 10 Cu NH

K



 

 

  

   

Strategy Because formation constants are typically very large, we begin by assuming that all the Cd2+_{ion is consumed and converted to complex ion. We}

then determine how much Cd2+_{is produced by the subsequent dissociation of the}

complex ion, a process for which the equilibrium constant is the reciprocal Kf.

In the presence of aqueous cyanide, cadmium(II) forms the complex ion Cd(CN)42-. Determine the molar concentration of free (uncomplexed)

cadmium(II) ion in solution when 0.20 mol of Cd(NO3)2 is dissolved in a liter of

2.0 M sodium cyanide (NaCN).

Solution From Table 17.5, the formation constant (Kf) for the complex ion

Cd(CN)42- is 7.1×1016. The reverse process,

Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)

has an equilibrium constant of 1/Kf = 1.4×10-17. The equilibrium expression for

the dissociation is

1.4×10-17 ₌[Cd2+][CN-]4

[Cd(CN)42-]

Worked Example 17.12 (cont.)

Solution The formation of complex ion will consume some of the cyanide originally present. Stoichiometry indicates that four CN-_{ions are required to react}

with one Cd2+_{ion. Therefore, the concentration of CN}-_{that we enter in the top}

row of the equilibrium table will be [2.0 M – 4(0.20 M)] = 1.2 M.

and, because the magnitude of K is so small, we can neglect x with respect to the initial concentrations of Cd(CN)42- and CN- (0.20 – x ≈ 0.20 and 1.2 + 4x ≈ 1.2),

so the solution becomes

and x = 1.4×10-18_M_.

Change in concentration (M) -x +x +4x

Equilibrium concentration (M) 0.20 - x x 1.2 + 4x

Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)

= 1.4×10 -17

[Cd2+_][CN-_]4

[Cd(CN)42-] x(1.2)4

0.20 =

Think About It When you assume that all the metal ion is consumed and converted to complex ion, it’s important to remember that some of the complexing agent (in this case, CN-_{ion) is}

consumed in the process. Don’t forget to adjust its concentration accordingly before entering it in the top row of the equilibrium table.

Separation of Ions Using Differences in Solubility

Some compounds can be separated based on fractional precipitation.

Fractional precipitation

is the separation of mixture based upon the

components’ solubilities.

17.6

Compound Ksp AgCl 1.6 x 10–10

AgBr 7.7 x 10–13

(12)

Strategy Silver nitrate dissociates in solution to give Ag+_{and NO} 3- ions.

Adding Ag+_{ions in sufficient amount will cause the slightly soluble ionic}

compounds AgCl and AgBr to precipitate from solution. Knowing the Ksp values

for AgCl and AgBr (and the concentrations of Cl-_{and Br}-_{already in solution), we}

can use the equilibrium expressions to calculate the maximum concentration of Ag+_{that can exist in solution without exceeding}_K

sp for each compound.

Silver nitrate is added slowly to a solution that is 0.020 M in Cl- ions and 0.020 M

in Br- _{ions. Calculate the concentration of Ag}+_{ions (in mol/L) required to initiate}

the precipitation of AgBr without precipitating AgCl.

Worked Example 17.13 (cont.)

Solution The solubility equilibria, Ksp values, and equilibrium expressions for

AgCl and AgBr are

AgCl(s) ⇌ Ag+₍_aq_{) + Cl}-₍_aq₎ _K_sp_{= 1.6}×10-10_{= [Ag}+_][Cl-_]

AgBr(s) ⇌ Ag+₍_aq_{) + Br}-₍_aq₎ _K_sp_{= 7.7}×10-13_{= [Ag}+_][Br-_]

Because the Ksp for AgBr is smaller (by a factor of more than 200), AgBr should

precipitate first; that is, it will require a lower concentration of added Ag+_{to begin}

precipitation. Therefore, we first solve for [Ag+_{] using the equilibrium expression}

for AgBr to determine the minimum Ag+_{concentration necessary to initiate}

precipitation of AgBr. We then solve for [Ag+_{] again, using the equilibrium}

expression for AgCl to determine the maximum Ag+_{concentration that can exist}

in the solution without initiating precipitation of AgCl.

Solving the AgBr equilibrium expression for Ag+_{concentration, we have} Ksp

[Br-_]

[Ag+_{] =} 7.7×10 -13

0.020

= = 3.9×10-11_M

Worked Example 17.13 (cont.)

Solution For AgBr to precipitate from solution, the silver ion concentration must exceed 3.9×10-11_M_{. Solving the AgCl equilibrium expression for the Ag}+

concentration, we have

For AgCl not to precipitate from solution, the silver ion concentration must stay below 8.0×10-9_M_{. Therefore, to precipitate the Br}-_{ions without precipitating the}

Cl-_{from this solution, the Ag}+_{concentration must be greater than 3.9}×10-11_M

and less than 8.0×10-9_M_. Ksp

[Cl-_]

[Ag+_{] =} 1.6×10 -10

0.020

= = 8.0×10-9 M

Think About It If we continue adding AgNO3 until the Ag+ concentration is

high enough to begin precipitation of AgCl, the concentration of Br-_{remaining in}

solution can also be determined using the Ksp expression.

Thus, by the time AgCl begins to precipitate, (9.6×10-5_M₎÷_(0.020M_{) = 0.0048,}

so less than 0.5 percent of the original bromide ion remains in the solution.

Ksp

[Ag+_]

[Br-_{] =} 7.7×10 -13

8.0×10 -9

= = 9.6×10-5 M

Qualitative analysis involves the

principle of selective precipitation

and can be used to identify the

types of ions present in a solution.

Separation of Ions Using Differences in Solubility

Key Concepts

17

Calculating the pH of a Buffer

Preparing a Buffer Solution with as Specific pH Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Strong Acid-Weak base Titrations Acid-Base Indicators

Solubility Product Expression and Ksp Calculations Involving Ksp and Solubility Predicting Precipitation Reactions The Common Ion Effect pH

Complex Ion Formation Fractional Precipitation