Chapter 17

Acid-Base Equilibria and Solubility Equilibria

 

17.5

a.

This is a weak acid problem.  Setting up the standard equilibrium table:

 

 

 

CH3COOH(aq) 

H+(aq)

+    CH3COO-(aq)

 

Initial (M):

0.40

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.40 - x)

 

x

x

 

 

 

x  =  [H+] = 2.7 ´ 10-3 M

 

pH  =  2.57

b.

In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving.

 

CH3COONa(aq)  ®  CH3COO-(aq) + Na+(aq)

 

Dissolving 0.20 M sodium acetate initially produces 0.20 M CH3COO- and 0.20 M Na+.  The sodium ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the equilibrium in part (a).

 

                                       

CH3COOH(aq) 

H+(aq)

+    CH3COO-(aq)

 

Initial (M):

0.40

 

0.00

0.20

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.40 - x)

 

x

(0.20 + x)

 

 

 

x  =  [H+] =  3.6 ´ 10-5 M

 

pH  =  4.44

 

Think About It:

Could you have predicted whether the pH should have increased or decreased after the addition of the sodium acetate to the pure 0.40 M acetic acid in part (a)?

An alternate way to work part (b) of this problem is to use the Henderson-Hasselbalch equation.

 

 

17.6

a.

This is a weak base problem.

                                           

                                       

 

 

Initial (M):

0.20

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.20 - x

 

 

+x

+x

 

 

 

 

x  =  1.9 ´ 10-3 M  =  [OH-]

 

pOH  =  2.72

 

pH  =  11.28

b.

The initial concentration of NH is 0.30 M from the salt NH4Cl.  We set up a table as in part (a).

 

 

                                       

 

 

Initial (M):

0.20

 

 

0.30

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.20 - x

 

 

0.30 + x

+x

                                           

 

 

x  =  1.2 ´ 10-5 M  =  [OH-]

 

pOH  =  4.92

 

pH  =  9.08

 

Alternatively, we could use the Henderson-Hasselbalch equation to solve this problem.  We get the value of Ka for the ammonium ion using the Kb for ammonia (Table 16.7) and Equation 16.8.  Substituting into the Henderson-Hasselbalch equation gives:

 

 

pH  =  9.25 - 0.18  =  9.07

 

Think About It:

Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its conjugate base and a weak base and its conjugate acid?

17.9

Strategy:

What constitutes a buffer system?  Which of the solutions described in the problem contains a weak acid and its salt (containing the weak conjugate base)?  Which contains a weak base and its salt (containing the weak conjugate acid)?  Why is the conjugate base of a strong acid not able to neutralize an added acid?

Solution:

The criteria for a buffer system are that we must have a weak acid and its salt (containing the weak conjugate base) or a weak base and its salt (containing the weak conjugate acid).

a.

HCl (hydrochloric acid) is a strong acid.  A buffer is a solution containing both a weak acid and a weak base.  Therefore, this is not a buffer system.

b.

H2SO4 (sulfuric acid) is a strong acid.  A buffer is a solution containing both a weak acid and a weak base.  Therefore, this is not a buffer system.

c.

This solution contains both a weak acid, H2PO and its conjugate base, HPO.  Therefore, this is a buffer system.

d.

HNO2 (nitrous acid) is a weak acid, and its conjugate base, NO (nitrite ion, the anion of the salt KNO2), is a weak base.  Therefore, this is a buffer system.

Only (c) and (d) are buffer systems.

17.10

a.

HCN is a weak acid, and its conjugate base, CN-, is a weak base.  Therefore, this is a buffer system.

b.

HSO is a weak acid, and its conjugate base, SO is a weak base.  Therefore, this is a buffer system.

c.

NH3 (ammonia) is a weak base, and its conjugate acid, NH is a weak acid.  Therefore, this is a buffer system.

d.

Because HI is a strong acid, its conjugate base, I-, is an extremely weak base.  This means that the I- ion will not combine with a H+ ion in solution to form HI.  Thus, this system cannot act as a buffer system.

17.11

Strategy:

The pH of a buffer system can be calculated using the Henderson-Hasselbalch equation (Equation 17.3).  The Ka of a conjugate acid such as NH is calculated using the Kb of its weak base (in this case, NH3) and Equation 16.8.

Solution:

NH(aq)    NH3(aq) + H+(aq)

 

Ka  =  Kw /1.8 ´ 10-5 = 5.56 ´ 10-10

 

pKa  = -log Ka = 9.26

 

pH = 8.89

17.12

a.

We summarize the concentrations of the species at equilibrium as follows:

 

                                                           

                                       

CH3COOH(aq) 

H+(aq)

+    CH3COO-(aq)

 

Initial (M):

2.0

 

0

2.0

Change (M):

­-x

 

+x

+x

Equilibrium (M):

2.0 - x

 

x

2.0 + x

 

 

 

Ka  =  [H+]

 

Taking the -log of both sides,

 

pKa  =  pH

 

Thus, for a buffer system in which the [weak acid] = [weak base],

 

pH  =  pKa

 

pH  =  -log(1.8 ´ 10-5)  =  4.74

b.

Similar to part (a),

 

pH  =  pKa  =  4.74

 

Buffer (a) will be a more effective buffer because the concentrations of acid and base components are ten times higher than those in (b).  Thus, buffer (a) can neutralize 10 times more added acid or base compared to buffer (b).

17.13

H2CO3(aq)    HCO(aq) + H+(aq)

 

 

 

 

 

 

 

17.14

Step 1:   Write the equilibrium that occurs between H2PO and HPO.  Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.

 

               

                                       

 

  

Initial (M):

0.15

 

0

0.10

Change (M):

­-x

 

+x

+x

Equilibrium (M):

0.15 - x

 

x

0.10 + x

 

 

Step 2:   Write the ionization constant expression in terms of the equilibrium concentrations.  Knowing the value of the equilibrium constant (Ka), solve for x.

 

 

You can look up the Ka value for dihydrogen phosphate (Ka2 for phosphoric acid) in Table 16.8 of your text.

 

 

 

x  =  [H+]  =  9.3 ´ 10-8 M

 

Step 3:   Having solved for the [H+], calculate the pH of the solution.

 

pH  =  -log[H+]  =  -log(9.3 ´ 10-8)  =  7.03

17.15

Using the Henderson-Hasselbalch equation:

 

 

Thus,

17.16

We can use the Henderson-Hasselbalch equation to calculate the ratio [HCO3-]/[H2CO3].  The Henderson-Hasselbalch equation is:

 

 

For the buffer system of interest, HCO3- is the conjugate base of the acid, H2CO3.  We can write:

 

 

 

The [conjugate base]/[acid] ratio is:

 

 

 

The buffer should be more effective against an added acid because ten times more base is present compared to acid.  Note that a pH of 7.40 is only a two significant figure number (Why?); the final result should only have two significant figures.

17.17

For the first part we use Ka for ammonium ion.  (Why?)  The Henderson-Hasselbalch equation is

 

 

For the second part, the acid-base reaction is

 

NH3(g) + H+(aq)  ®  NH(aq)

 

We find the number of moles of HCl added

 

 

The number of moles of NH3 and NH4+ originally present are

 

 

Using the acid-base reaction, we find the number of moles of NH3 and NH after addition of the HCl.

 

 

                                       

 

Initial (mol):

0.013

0.0010

 

0.013

Change (mol):

-0.0010

-0.0010

 

+0.0010

Final (mol):

0.012

0

 

0.014

 

 

 

 

 

 

 

We find the new pH:

 

17.18

As calculated in Problem 17.12, the pH of this buffer system is equal to pKa.

 

                                                              pH  =  pKa  =  -log(1.8 ´ 10-5)  =  4.74

 

a.

The added NaOH will react completely with the acid component of the buffer, CH3COOH.  NaOH ionizes completely; therefore, 0.080 mol of OH- are added to the buffer.

 

Step 1:   The neutralization reaction is:

 

                                       

CH3COOH(aq)

+    OH-(aq)

CH3COO-(aq)

+     H2O(l)

Initial (mol):

1.00

0.080

 

1.00

 

Change (mol):

-0.080

-0.080

 

+0.080

 

Final (mol):

0.92

0

 

1.08

 

 

 

Step 2:   Now, the acetic acid equilibrium is reestablished.  Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.

               

                                       

CH3COOH(aq) 

H+(aq)

+    CH3COO-(aq)

 

Initial (M):

0.92

 

0

1.08

Change (M):

­-x

 

+x

+x

Equilibrium (M):

0.92 - x

 

x

1.08 + x

 

Write the Ka expression, then solve for x.

 

 

 

x  =  [H+]  =  1.5 ´ 10-5 M

 

Step 3:   Having solved for the [H+], calculate the pH of the solution.

 

pH  =  -log[H+]  =  -log(1.5 ´ 10-5)  =  4.82

 

The pH of the buffer increased from 4.74 to 4.82 upon addition of 0.080 mol of strong base.

b.

The added acid will react completely with the base component of the buffer, CH3COO-.  HCl ionizes completely; therefore, 0.12 mol of H+ ion are added to the buffer

 

Step 1:   The neutralization reaction is:

 

 

                                       

CH3COO-(aq)

+    H+(aq)  

CH3COOH(aq)

 

Initial (mol):

1.00

0.12

 

1.00

Change (mol):

-0.12

-0.12

 

+0.12

Final (mol):

0.88

0

 

1.12

 

 

 

 

 

               

 

Step 2:   Now, the acetic acid equilibrium is reestablished.  Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.

 

               

Step 2:   Now, the acetic acid equilibrium is reestablished.  Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.

 

 

                                       

CH3COOH(aq) 

H+(aq)

+    CH3COO-(aq)

 

Initial (M):

1.12

 

0

0.88

Change (M):

­-x

 

+x

+x

Equilibrium (M):

1.12 - x

 

x

0.88 + x

 

               

Write the Ka expression, then solve for x.

 

 

 

x  =  [H+]  =  2.3 ´ 10-5 M

 

Step 3:   Having solved for the [H+], calculate the pH of the solution.

 

pH  =  -log[H+]  =  -log(2.3 ´ 10-5)  =  4.64

 

The pH of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid.

17.19

In order for the buffer solution to function effectively, the pKa of the acid component must be close to the desired pH.   We write

 

                                                                       

 

                                                                      

 

Therefore, the proper buffer system is Na2A/NaHA.

17.20

For a buffer to function effectively, the concentration of the acid component must be roughly equal to the conjugate base component.  According to Equation 17.3 of the text, this occurs when the desired pH is close to the pKa of the acid, that is, when pH » pKa,

 

or

 

To prepare a solution of a desired pH, we should choose a weak acid with a pKa value close to the desired pH.  Calculating the pKa for each acid:

 

                                                            For HA,                pKa  =  -log(2.7 ´ 10-3)  =  2.57

 

                                                            For HB,                pKa  =  -log(4.4 ´ 10-6)  =  5.36

 

                                                            For HC,                pKa  =  -log(2.6 ´ 10-9)  =  8.59

 

 

The buffer solution with a pKa closest to the desired pH is HC.  Thus, HC is the best choice to prepare a buffer solution with pH  =  8.60.

17.21

1.

In order to function as a buffer, a solution must contain species that will consume both acid and base. Solution (a) contains HA-, which can consume either acid or base:

 

HA-  +  H+  ®  H2A

 

HA-  +  OH-  ®  A2-  +  H2O

 

and A2-, which can consume acid:

 

A2-  +  H+  ®  HA-

 

Solution (b) contains HA-, which can consume either acid or base, and H2A, which can consume base:

 

H2A  +  OH-  ®  HA-  +  H2O

 

Solution (c) contains only H2A and consequently can consume base but not acid.  Solution (c) cannot function as a buffer.

 

Solution (d) contains HA- and A2-.

 

Solutions (a), (b), and (c) can function as buffers.

2.

Solution (a) should be the most effective buffer because it has the highest concentrations of acid- and base-consuming species. 

17.22

1.

Solution (d) has the lowest pH because it has the highest ratio of HA to A- (6:4).  Solution (a) has the highest pH because it has the lowest ratio of HA to A- (3:5). 

2.

After addition of two H+ ions to solution (a), there will be two species present: HA and A-. 

3.

After addition of two OH- ions to solution (b), there will be two species present: HA and A-.

17.27

Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid.

 

 

17.28

We want to calculate the molar mass of the diprotic acid.  The mass of the acid is given in the problem, so we need to find moles of acid in order to calculate its molar mass.

 

                                               

 

The neutralization reaction is:

 

2KOH(aq) + H2A(aq)    K2A(aq) + 2H2O(l)

 

From the volume and molarity of the base needed to neutralize the acid, we can calculate the number of moles of H2A reacted.

 

 

We know that 0.500 g of the diprotic acid were reacted (1/10 of the 250 mL was tested).  Divide the number of grams by the number of moles to calculate the molar mass.

 

 

17.29

The neutralization reaction is:

 

H2SO4(aq) + 2NaOH(aq)  ®  Na2SO4(aq) + 2H2O(l)

 

Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write:

 

 

17.30

We want to calculate the molarity of the Ba(OH)2 solution.  The volume of the solution is given (19.3 mL), so we need to find the moles of Ba(OH)2 to calculate the molarity.

 

                                               

 

The neutralization reaction is:

 

2HCOOH + Ba(OH)2  ®  (HCOO)2Ba + 2H2O

 

From the volume and molarity of HCOOH needed to neutralize Ba(OH)2, we can determine the moles of Ba(OH)2 reacted.

 

 

The molarity of the Ba(OH)2 solution is:

 

17.31

a.

Since the acid is monoprotic, the moles of acid equals the moles of base added.

 

HA(aq) + NaOH(aq)    NaA(aq) + H2O(l)

 

 

We know the mass of the unknown acid in grams and the number of moles of the unknown acid.

 

b.

The number of moles of NaOH in 10.0 mL of solution is

 

 

The neutralization reaction is:

 

 

                                       

HA(aq)     

+   NaOH(aq)

NaA(aq) 

+     H2O(l)

Initial (mol):

0.00116

6.33 ´ 10-4

 

0

 

Change (mol):

-6.33 ´ 10-4

-6.33 ´ 10-4

 

+6.33 ´ 10-4

 

Final (mol):

5.3 ´ 10-4

0

 

6.33 ´ 10-4

 

 

Now, the weak acid equilibrium will be reestablished.  The total volume of solution is 35.0 mL.

 

 

 

We can calculate the [H+] from the pH.

 

[H+]  =  10-pH  =  10-5.87  =  1.35 ´ 10-6 M

 

                         

                                       

HA(aq)       

   H+(aq)    +

  A-(aq)

Initial (M):

0.015

 

0

0.0181

Change (M):

­-1.35 ´ 10-6

 

+1.35 ´ 10-6

+1.35 ´ 10-6

Equilibrium (M):

0.015

 

1.35 ´ 10-6

0.0181

 

Substitute the equilibrium concentrations into the equilibrium constant expression to solve for Ka.

 

 

17.32

The resulting solution is not a buffer system.  There is excess NaOH and the neutralization is well past the equivalence point.

 

 

 

 

                                       

CH3COOH(aq)

+    NaOH(aq)   

CH3COONa(aq)

+   H2O(l)

Initial (mol):

0.0500

0.0835

 

0

 

Change (mol):

-0.0500

-0.0500

 

+0.0500

 

Final (mol):

0

0.0335

 

0.0500

 

 

The volume of the resulting solution is 1.00 L (500 mL + 500 mL = 1000 mL).

 

 

 

 

 

                                       

                                       

CH3COO-(aq)

+     H2O(l) 

CH3COOH(aq)

+   OH-(aq)

 

Initial (M):

0.0500

 

 

0

0.0335

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.0500 - x

 

 

x

0.0335 + x

 

 

 

x  =  [CH3COOH]  =  8.4 ´ 10-10 M

17.33

HCl(aq) + CH3NH2(aq)    CH3NH(aq)  +  Cl-(aq)

 

Since the concentrations of acid and base are equal, equal volumes of each solution will need to be added to reach the equivalence point.  Therefore, the solution volume is doubled at the equivalence point, and the concentration of the conjugate acid from the salt, CH3NH3+, is:

 

 

The conjugate acid undergoes hydrolysis.

 

 

                                       

 

Initial (M):

0.10

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.10 - x

 

 

x

x

 

 

 

Assuming that, 0.10 - x » 0.10

 

x  =  [H3O+]  =  1.5 ´ 10-6 M

 

pH  =  5.82

17.34

Let's assume we react 1 L of HCOOH with 1 L of NaOH.

 

                                       

                                       

HCOOH(aq)

+    NaOH(aq) 

HCOONa(aq)

+     H2O(l)

 

Initial (mol):

0.10

0.10

 

0

 

Change (mol):

-0.10

-0.10

 

+0.10

 

Final (mol):

0

0

 

0.10

 

 

                                                 

 

 

 

 

 

The solution volume has doubled (1 L + 1 L = 2 L).  The concentration of HCOONa is:

 

 

HCOO-(aq) is a weak base.  The hydrolysis is:

 

                                       

                                       

HCOO-(aq)

+    H2O(l) 

HCOOH(aq)

+    OH-(aq)

 

Initial (M):

0.050

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.050 - x

 

 

x

x

 

 

 

x  =  1.7 ´ 10-6 M  =  [OH-]

 

pOH  =  5.77

 

pH  =  8.23

17.35

The reaction between CH3COOH and KOH is:

 

CH3COOH(aq) + KOH(aq)  ®  CH3COOK(aq) + H2O(l)

 

We see that 1 mole CH3COOH is stoichiometrically equivalent to 1 mol KOH.  Therefore, at every stage of titration, we can calculate the number of moles of acid reacting with base, and the pH of the solution is determined by the excess acid or base left over.  At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is CH3COOK.

a.

No KOH has been added.  This is a weak acid calculation.

                                       

                                       

CH3COOH(aq)

+     H2O(l) 

H3O+(aq)

+    CH3COO-(aq)

Initial (M):

0.100

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.100 - x

 

 

x

x

 

 

 

x  =  1.34 ´ 10-3 M  =  [H3O+]

 

pH  =  2.87

b.

The number of moles of CH3COOH originally present in 25.0 mL of solution is:

 

 

The number of moles of KOH in 5.0 mL is:

 

 

We work with moles at this point because when two solutions are mixed, the solution volume increases.  As the solution volume increases, molarity will change, but the number of moles will remain the same.  The changes in number of moles are summarized.

 

                                                                 

                                                                 

CH3COOH(aq)     

+    KOH(aq)

CH3COOK(aq)

+   H2O(l)

Initial (mol):

2.50 ´ 10-3

1.00 ´ 10-3

 

0

 

Change (mol):

-1.00 ´ 10-3

-1.00 ´ 10-3

 

+1.00 ´ 10-3

 

Final (mol):

1.50 ´ 10-3

0

 

1.00 ´ 10-3

 

 

At this stage, we have a buffer system made up of CH3COOH and CH3COO- (from the salt, CH3COOK).  We use the Henderson-Hasselbalch equation to calculate the pH.

 

 

 

pH  =  4.56

c.

This part is solved similarly to part (b).

 

The number of moles of KOH in 10.0 mL is:

 

 

The changes in number of moles are summarized.

 

                                                                 

                                                                 

CH3COOH(aq)     

+   KOH(aq)

CH3COOK(aq)

+    H2O(l)

Initial (mol):

2.50 ´ 10-3

2.00 ´ 10-3

 

0

 

Change (mol):

-2.00 ´ 10-3

-2.00 ´ 10-3

 

+2.00 ´ 10-3

 

Final (mol):

0.50 ´ 10-3

0

 

2.00 ´ 10-3

 

 

At this stage, we have a buffer system made up of CH3COOH and CH3COO- (from the salt, CH3COOK).  We use the Henderson-Hasselbalch equation to calculate the pH.

 

 

 

pH  =  5.34

d.

We have reached the equivalence point of the titration.  2.50 ´ 10-3 mole of CH3COOH reacts with
2.50 ´ 10-3 mole KOH to produce 2.50 ´ 10-3 mole of CH3COOK.  The only major species present in solution at the equivalence point is the salt, CH3COOK, which contains the conjugate base, CH3COO-.  Let's calculate the molarity of CH3COO-.  The volume of the solution is:  (25.0 mL + 12.5 mL = 37.5 mL = 0.0375 L).

 

 

We set up the hydrolysis of CH3COO-, which is a weak base.

 

                                       

                                       

CH3COO-(aq)

+   H2O(l) 

CH3COOH(aq)

+      OH-(aq)

 

Initial (M):

0.0667

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.0667 - x

 

 

x

x

 

 

 

 

x  =  6.09 ´ 10-6 M  =  [OH-]

 

pOH  =  5.22

 

pH  =  8.78

e.

We have passed the equivalence point of the titration.  The excess strong base, KOH, will determine the pH at this point.  The moles of KOH in 15.0 mL are:

 

 

The changes in number of moles are summarized.

 

                                                                 

                                                                 

CH3COOH(aq)     

+   KOH(aq)

CH3COOK(aq)

+  H2O(l)

Initial (mol):

2.50 ´ 10-3

3.00 ´ 10-3

 

0

 

Change (mol):

-2.50 ´ 10-3

-2.50 ´ 10-3

 

+2.50 ´ 10-3

 

Final (mol):

0

0.50 ´ 10-3

 

2.50 ´ 10-3

 

 

 

Let's calculate the molarity of the KOH in solution.  The volume of the solution is now 40.0 mL =

0.0400 L.

 

 

KOH is a strong base.  The pOH is:

 

pOH  =  -log(0.0125)  =  1.90

 

pH  =  12.10

17.36

The reaction between NH3 and HCl is:

 

NH3(aq) + HCl(aq)  ®  NH4Cl(aq)

 

We see that 1 mole NH3 is stoichiometrically equivalent to 1 mol HCl.  Therefore, at every stage of titration, we can calculate the number of moles of base reacting with acid, and the pH of the solution is determined by the excess base or acid left over.  At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is NH4Cl.

a.

No HCl has been added.  This is a weak base problem.

 

 

 

 

x  =  2.3 ´ 10-3 M  =  [OH-]

 

pOH  =  2.64

 

pH  =  11.36

b.

The number of moles of NH3 originally present in 10.0 mL of solution is:

 

 

The number of moles of HCl in 10.0 mL is:

 

 

We work with moles at this point because when two solutions are mixed, the solution volume increases.  As the solution volume increases, molarity will change, but the number of moles will remain the same.  The changes in number of moles are summarized.

 

 

                                       

NH3(aq)    +     

HCl(aq)

NH4Cl(aq)

Initial (mol):

3.00 ´ 10-3

1.00 ´ 10-3

 

0

Change (mol):

-1.00 ´ 10-3

-1.00 ´ 10-3

 

+1.00 ´ 10-3

Final (mol):

2.00 ´ 10-3

0

 

1.00 ´ 10-3

 

 

 

 

 

                                                                  NH3(aq)      +      HCl(aq)      ®      NH4Cl(aq)

               

At this stage, we have a buffer system made up of NH3 and NH (from the salt, NH4Cl).  We use the Henderson-Hasselbalch equation to calculate the pH.

 

 

 

pH  =  9.55

c.

This part is solved similarly to part (b).

 

                          The number of moles of HCl in 20.0 mL is:

 

                                           

 

The changes in number of moles are summarized.

 

 

                                       

NH3(aq)      +

  HCl(aq)     

NH4Cl(aq)

Initial (mol):

3.00 ´ 10-3

2.00 ´ 10-3

 

0

Change (mol):

-2.00 ´ 10-3

-2.00 ´ 10-3

 

+2.00 ´ 10-3

Final (mol):

1.00 ´ 10-3

0

 

2.00 ´ 10-3

 

 

 

 

 

 

 

At this stage, we have a buffer system made up of NH3 and NH (from the salt, NH4Cl).  We use the Henderson-Hasselbalch equation to calculate the pH.

 

 

 

pH  =  8.95

d.

We have reached the equivalence point of the titration.  3.00 ´ 10-3 mole of NH3 reacts with
3.00 ´ 10-3 mole HCl to produce 3.00 ´ 10-3 mole of NH4Cl.  The only major species present in solution at the equivalence point is the salt, NH4Cl, which contains the conjugate acid, NH.  Let's calculate the molarity of NH.  The volume of the solution is:  (10.0 mL + 30.0 mL = 40.0 mL = 0.0400 L).

 

We set up the hydrolysis of NH, which is a weak acid.

 

 

 

                                       

Initial (M):

0.0750

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.0750 - x

 

 

x

x

 

 

 

x  =  6.5 ´ 10-6 M  =  [H3O+]

 

pH  =  5.19

e.

We have passed the equivalence point of the titration.  The excess strong acid, HCl, will determine the pH at this point.  The moles of HCl in 40.0 mL are:

 

 

The changes in number of moles are summarized.

 

                                       

NH3(aq)    +

 HCl(aq)     

NH4Cl(aq)

Initial (mol):

3.00 ´ 10-3

4.00 ´ 10-3

 

0

Change (mol):

-3.00 ´ 10-3

-3.00 ´ 10-3

 

+3.00 ´ 10-3

Final (mol):

0

1.00 ´ 10-3

 

3.00 ´ 10-3

 

 

 

 

 

 

 

Let's calculate the molarity of the HCl in solution.  The volume of the solution is now                                 50.0 mL = 0.0500 L.

 

 

HCl is a strong acid.  The pH is:

 

pH  =  -log(0.0200)  =  1.70

17.37

a.

HCOOH is a weak acid and NaOH is a strong base.  Suitable indicators are cresol red and phenolphthalein.

b.

HCl is a strong acid and KOH is a strong base.  Most of the indicators in Table 17.3 are suitable for a strong acid-strong base titration.  Exceptions are thymol blue and, to a lesser extent, bromophenol blue and methyl orange.

c.

HNO3 is a strong acid and CH3NH2 is a weak base.  Suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue.

17.38

CO2 in the air dissolves in the solution:

 

CO2 + H2O   H2CO3

 

The carbonic acid neutralizes the NaOH, lowering the pH.  When the pH is lowered, the phenolphthalein indicator returns to its colorless form.

17.39

The weak acid equilibrium is

 

HIn(aq)    H+(aq)  +  In-(aq)

 

We can write a Ka expression for this equilibrium.

 

Rearranging,

 

From the pH, we can calculate the H+ concentration.

 

[H+]  =  10-pH  =  10-4  =  1.0 ´ 10-4 M

 

 

Since the concentration of HIn is 100 times greater than the concentration of In-, the color of the solution will be that of HIn, the nonionized formed.  The color of the solution will be red.

17.40

When [HIn] » [In-] the indicator color is a mixture of the colors of HIn and In-.  In other words, the indicator color changes at this point.  When [HIn] » [In-] we can write:

 

 

[H+]  =  Ka  =  2.0 ´ 10-6

 

pH  =  5.70

17.41

1.

Diagram (c)

2.

Diagram (b)

3.

Diagram (d)

4.

Diagram (a).  The pH at the equivalence point is below 7 (acidic) because the species in solution at the equivalence point is the conjugate acid of a weak base.

17.42

1.

Diagram (b)

2.

Diagram (d)

3.

Diagram (a)

4.

Diagram (c).  The pH at the equivalence point is above 7 (basic) because the species in solution at the equivalence point is the conjugate base of a weak acid.

17.49

a.

The solubility equilibrium is given by the equation

 

AgI(s)   Ag+(aq) + I-(aq)

 

The expression for Ksp is given by

 

Ksp   =  [Ag+][I-]

 

The value of Ksp can be found in Table 17.4 of the text.  If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be

 

b.

The value of Ksp for aluminum hydroxide can be found in Table 17.4 of the text.  The equilibrium expressions are:

 

Al(OH)3(s)   Al3+(aq) + 3OH-(aq)

 

Ksp  =  [Al3+][OH-]3

 

Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is:

 

Think About It:

What is the pH of this solution?  Will the aluminum concentration change if the pH is altered?

17.50

In each part, we can calculate the number of moles of compound dissolved in one liter of solution
(the molar solubility).  Then, from the molar solubility, s, we can determine Ksp.

a.

 

Consider the dissociation of SrF2 in water.  Let s be the molar solubility of SrF2.

 

                                       

SrF2(s)   

Sr2+(aq)  

+     2F-(aq)

 

Initial (M):

 

 

0

0

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

 

 

s

2s

 

Ksp  =  [Sr2+][F-]2  =  (s)(2s)2  =  4s3

 

The molar solubility (s) was calculated above.  Substitute into the equilibrium constant expression to solve for Ksp.

 

Ksp  =  [Sr2+][F-]2  =  4s3  =  4(5.8 ´ 10-4)3  =  7.8 ´ 10-10

b.

 

(b) is solved in a similar manner to (a)

 

The equilibrium equation is:

 

                                       

 

 

Initial (M):

 

 

0

0

Change (M):

-s

 

+3s

+s

Equilibrium (M):

 

 

3s

s

 

 

 

 

 

 

 

 

Ksp  =  [Ag+]3[PO43-]  =  (3s)3(s)  =  27s4  =  27(1.6 ´ 10-5)4  =  1.8 ´ 10-18

17.51

For MnCO3 dissolving, we write

 

MnCO3(s)   Mn2+(aq) + CO(aq)

 

For every mole of MnCO3 that dissolves, one mole of Mn2+ will be produced and one mole of CO will be produced.  If the molar solubility of MnCO3 is s mol/L, then the concentrations of Mn2+ and CO are:

 

[Mn2+]  =  [CO]  =  s  =  4.2 ´ 10-6 M

 

Ksp  =  [Mn2+][CO]  =  s2  =  (4.2 ´ 10-6)2  =  1.8 ´ 10-11

17.52

First, we can convert the solubility of MX in g/L to mol/L.

 

 

The equilibrium equation is:

 

                                       

MX(s)   

Mn+(aq)  

+      Xn-(aq)

 

Initial (M):

 

 

0

0

Change (M):

­-s

 

+s

+s

Equilibrium (M):

 

 

s

s

 

 

Ksp  =  [Mn+][Xn-]  =  s2  =  (1.34 ´ 10-5)2  =  1.80 ´ 10-10

17.53

The charges of the M and X ions are +3 and -2, respectively (are other values possible?).  We first calculate the number of moles of M2X3 that dissolve in 1.0 L of water.  We carry an additional significant figure throughout this calculation to minimize rounding errors.

 


 

The molar solubility, s, of the compound is therefore 1.3 ´ 10-19 M.  At equilibrium the concentration of M3+ must be 2s and that of X2- must be 3s.

 

Ksp  =  [M3+]2[X2-]3  =  [2s]2[3s]3  =  108s5

 

Since these are equilibrium concentrations, the value of Ksp can be found by simple substitution

 

Ksp  =  108s5  =  108(1.25 ´ 10-19)5  =  3.3 ´ 10-93

17.54

We can look up the Ksp value of CaF2 in Table 17.4 of the text.  Then, setting up the dissociation equilibrium of CaF2 in water, we can solve for the molar solubility, s.

 

Consider the dissociation of CaF2 in water.

 

 

                                       

CaF2(s)     

Ca2+(aq)    

+     2F-(aq)

 

Initial (M):

 

 

0

0

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

 

 

s

2s

 

 

Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression.  Therefore, the concentration of CaF2 is not important.

 

            Substitute the value of Ksp and the concentrations of Ca2+ and F- in terms of s into the solubility product expression to solve for s, the molar solubility.

 

Ksp  =   [Ca2+][F-]2

 

4.0 ´ 10-11  =  (s)(2s)2

 

4.0 ´ 10-11  =  4s3

 

s  =  molar solubility  =  2.2 ´ 10-4 mol/L

 

The molar solubility indicates that 2.2 ´ 10-4 mol of CaF2 will dissolve in 1 L of an aqueous solution.

17.55

Let s be the molar solubility of Zn(OH)2.  The equilibrium concentrations of the ions are then

 

[Zn2+] = s and [OH-] = 2s

 

Ksp  =  [Zn2+][OH-]2  =  (s)(2s)2  =  4s3  =  1.8 ´ 10-14

 

s =

 

[OH-] = 2s = 3.30 ´ 10-5 M and pOH = 4.48

 

pH  =  14.00 - 4.48  =  9.52

Think About It:

If the Ksp of Zn(OH)2 were smaller by many more powers of ten, would 2s still be the hydroxide ion concentration in the solution?

17.56

First we can calculate the OH- concentration from the pH.

 

pOH  =  14.00 - pH

 

pOH  =  14.00 - 9.68  =  4.32

 

[OH-]  =  10-pOH  =  10-4.32  =  4.8 ´ 10-5 M

 

The equilibrium equation is:

 

MOH(s)   M+(aq)  +  OH-(aq)

 

From the balanced equation we know that [M+] = [OH-]

 

Ksp  =  [M+][OH-]  =  (4.8 ´ 10-5)2  =  2.3 ´ 10-9

17.57

According to the solubility rules, the only precipitate that might form is BaCO3.

 

Ba2+(aq) + CO(aq)  ®  BaCO3(s)

 

The number of moles of Ba2+ present in the original 20.0 mL of Ba(NO3)2 solution is

 

 

The total volume after combining the two solutions is 70.0 mL.  The concentration of Ba2+ in 70 mL is

 

 

The number of moles of CO present in the original 50.0 mL Na2CO3 solution is

 

 

The concentration of CO in the 70.0 mL of combined solution is

 

 

Now we must compare Q and Ksp. From Table 17.4 of the text, the Ksp for BaCO3 is 8.1 ´ 10-9.  As for Q,

 

Q  =  [Ba2+]0[CO]0  =  (2.9 ´ 10-2)(7.1 ´ 10-2)  =  2.1 ´ 10-3

 

Since (2.1 ´ 10-3) > (8.1 ´ 10-9), then Q > Ksp.  Therefore, yes, BaCO3 will precipitate.

17.58

The net ionic equation is:

 

Sr2+(aq) + 2F-(aq)    SrF2(s)

 

Let’s find the limiting reagent in the precipitation reaction.

 

 

From the stoichiometry of the balanced equation, twice as many moles of F- are required to react with Sr2+.  This would require 0.0076 mol of F-, but we only have 0.0045 mol.  Thus, F- is the limiting reagent.

 

Let’s assume that the above reaction goes to completion.  Then, we will consider the equilibrium that is established when SrF2 partially dissociates into ions.

 

 

                                       

Sr2+(aq)   

+    2 F-(aq) 

SrF2(s)

Initial (mol):

0.0038

0.0045

 

0

Change (mol):

-0.00225

-0.0045

 

+0.00225

Final (mol):

0.00155

0

 

0.00225

 

 

 

 

 

 

 

 

Now, let’s establish the equilibrium reaction.  The total volume of the solution is 100 mL = 0.100 L.  Divide the above moles by 0.100 L to convert to molar concentration.

 

 

                                       

SrF2(s)   

Sr2+(aq)  

+    2F-(aq)

 

Initial (M):

0.0225

 

0.0155

0

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

0.0225 - s

 

0.0155 + s

2s

 

 

Write the solubility product expression, then solve for s.

 

Ksp  =  [Sr2+][F-]2

 

2.0 ´ 10-10  =  (0.0155 + s)(2s)2  »  (0.0155)(2s)2

 

s  =  5.7 ´ 10-5 M

 

[F-]  =  2s  =  1.1 ´ 10-4 M

 

[Sr2+]  =  0.0155 + s  =  0.016 M

 

Both sodium ions and nitrate ions are spectator ions and therefore do not enter into the precipitation reaction.

 

 

17.59

Strategy:

The addition of soluble salts to the solution changes the concentrations of M2+ and A–.  The salt which causes the largest change in concentration (and therefore a change in the reaction quotient, Q) will cause the precipitation of the greatest quantity of MA2.

Setup:

The balanced equation is

 

MA2    M2+ + 2A–

 

Therefore,

 

Ksp = [M2+][A–]2

Solution:

Since the anion, A–, in the Ksp equation is squared, increasing its value will have the greater impact on the Q value than an equivalent change in the cation, M2+. 

 

AlA3 contains three anions, whereas NaA and BaA2, contain only one or two.  M3(PO4)2 contains three cations, while M(NO3)2 contains only one cation.  Since the [A–] is squared in the Ksp equation, changing its concentration will have a greater impact on the precipitation than the three-fold change in [M2+].  Therefore, AlA3 will cause the greatest quantity of  MA2 to precipitate.

17.60

Strategy:

If the reaction quotient, Q, is less than or equal to Ksp, then no precipitate will form. 

Setup:

The equation for the dissociation is:

 

M2B   2M+ + B–

 

Therefore,

 

Ksp = [M+]2[B–]

 

The saturated solution is at equilibrium.  There are six cations and three anions.  This gives:

 

Ksp = [M+]2[B–] = (6)2(3) = 108

 

Since the volumes are equal and additive when combined, the volume doubles and the concentration of each ion is reduced by one-half. 

Solution:

a.

Diagram 2 has eight cations and diagram (a) has fourteen anions.  When added to an equal volume of solution, the concentration is reduced by one-half, yielding concentrations of four and seven, respectively.

 

This gives:

 

Q = [M+]2[B–] = (4)2(7) = 112

 

Since 112 is greater than 108, a precipitate will form.

b.

Diagram (b) has twelve anions.  The number of cations was determined in part (a).  This gives:

 

Q = [M+]2[B–] = (4)2(6) = 96

 

Since 96 is less than 108, no precipitate will form.

c.

Diagram (c) has four anions.  This gives:

 

Q = [M+]2[B–] = (4)2(2) = 32

 

Since 32 is less than 108, no precipitate will form.

d.

Diagram (d) has eight anions.  This gives:

 

Q = [M+]2[B–] = (4)2(4) = 64

 

Since 64 is less than 108, no precipitate will form.

e.

Diagram (e) has ten anions.   This gives:

 

Q = [M+]2[B–] = (4)2(5) = 80

 

Since 80 is less than 108, no precipitate will form.

Solutions (b), (c), (d) and (e) can be added to the solution of MNO3 without causing a precipitate to form.

17.64

First let s be the molar solubility of CaCO3 in this solution.

 

 

                                       

 

 

Initial (M):

 

 

0.050

0

Change (M):

-s

 

+s

+s

Equilibrium (M):

 

 

(0.050 + s)

s

 

 

                                                        CaCO3(s)        Ca2+(aq)   +  CO(aq)

               

                                                        CaCO3(s)        Ca2+(aq)   +  CO(aq)

               

 

Ksp  =  [Ca2+][CO]  =  (0.050 + s)s  =  8.7 ´ 10-9

 

We can assume 0.050 + s » 0.050, then

 

 

The mass of CaCO3 can then be found.

 

17.65

Strategy:

In parts (b) and (c), this is a common-ion problem.  In part (b), the common ion is Br-, which is supplied by both PbBr2 and KBr.  Remember that the presence of a common ion will affect only the solubility of PbBr2, but not the Ksp value because it is an equilibrium constant.  In part (c), the common ion is Pb2+, which is supplied by both PbBr2 and Pb(NO3)2.

Solution:

a.

Set up a table to find the equilibrium concentrations in pure water.

 

                                       

PbBr2(s)     

Pb2+(aq)   

+   2Br-(aq)

Initial (M):

 

 

0

0

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

 

 

s

2s

 

 

Ksp  =  [Pb2+][Br-]2

 

8.9 ´ 10-6  =  (s)(2s)2

 

s  =  molar solubility  =  0.013 M or 1.3 ´ 10-2 M

b.

Set up a table to find the equilibrium concentrations in 0.20 M KBr.  KBr is a soluble salt that ionizes completely giving an initial concentration of Br- = 0.20 M.

 

                                       

PbBr2(s)     

Pb2+(aq)   

+   2Br-(aq)

Initial (M):

 

 

0

0.20

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

 

 

s

0.20 + 2s

 

Ksp  =  [Pb2+][Br-]2

 

8.9 ´ 10-6  =  (s)(0.20 + 2s)2

 

8.9 ´ 10-6  »  (s)(0.20)2

 

s  =  molar solubility  =  2.2 ´ 10-4 M

 

Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 2.2 ´ 10-4 M as a result of the common ion (Br-) effect.

c.

Set up a table to find the equilibrium concentrations in 0.20 M Pb(NO3)2.  Pb(NO3)2 is a soluble salt that dissociates completely giving an initial concentration of [Pb2+] = 0.20 M.

                                                 

                                       

PbBr2(s)     

Pb2+(aq)   

+   2Br-(aq)

Initial (M):

0.20

 

0

 

Change (M):

­-s

 

+s

+2s

Equilibrium (M):

 

 

0.20 + s

2s

 

 

Ksp  =  [Pb2+][Br-]2

 

8.9 ´ 10-6  =  (0.20 + s)(2s)2

 

8.9 ´ 10-6  »  (0.20)(2s)2

 

s  =  molar solubility  =  3.3 ´ 10-3 M

 

Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 3.3 ´ 10-3 M as a result of the common ion (Pb2+) effect.

 

Think About It:

You should also be able to predict the decrease in solubility due to a common-ion using
Le Châtelier's principle.  Adding Br- or Pb2+ ions shifts the system to the left, thus decreasing the solubility of PbBr2.

17.66

We first calculate the concentration of chloride ion in the solution.

 

 

                                       

AgCl(s)  

Ag+(aq)    

+   Cl-(aq)

 

Initial (M):

 

 

0.000

0.180

Change (M):

­-s

 

+s

+s

Equilibrium (M):

 

 

s

(0.180 + s)

 

If we assume that (0.180 + s) » 0.180, then

 

Ksp  =  [Ag+][Cl-]  =  1.6 ´ 10-10

 

 

The molar solubility of AgCl is 8.9 ´ 10-10 M.

17.67

a.

The equilibrium equation is:

 

 

                                       

 

 

Initial (M):

 

 

0

0

Change (M):

-s

 

+s

+s

Equilibrium (M):

 

 

s

s

 

 

 

 

Ksp  =  [Ba2+][SO]

 

1.1 ´ 10-10  =  s2

 

s  =  1.0 ´ 10-5 M

 

The molar solubility of BaSO4 in pure water is 1.0 ´ 10-5 mol/L.

b.

The initial concentration of SO is 1.0 M.  

 

                                       

 

 

Initial (M):

 

 

0

1.0

Change (M):

-s

 

+s

+s

Equilibrium (M):

 

 

s

1.0 + s

 

 

 

 

 

 

 

 

Ksp  =  [Ba2+][SO]

 

1.1 ´ 10-10  =  (s)(1.0 + s)  »  (s)(1.0)

 

s  =  1.1 ´ 10-10 M

 

Due to the  common ion effect, the molar solubility of BaSO4 decreases to 1.1 ´ 10-10 mol/L in
1.0 M SO(aq) compared to 1.0 ´ 10-5 mol/L in pure water.

17.68

When the anion of a salt is a base, the salt will be more soluble in acidic solution because the hydrogen ion decreases the concentration of the anion (Le Chatelier's principle):

 

B-(aq) + H+(aq)  HB(aq)

a.

BaSO4 will be slightly more soluble because SO is a base (although a weak one).

b.

The solubility of PbCl2 in acid is unchanged over the solubility in pure water because HCl is a strong acid, and therefore Cl- is a negligibly weak base.

c.

Fe(OH)3 will be more soluble in acid because OH- is a base.

d.

CaCO3 will be more soluble in acidic solution because the CO ions react with H+ ions to form CO2 and H2O.  The CO2 escapes from the solution, shifting the equilibrium.

17.69

a.

I- is the conjugate base of the strong acid HI

b.

SO (aq) is a weak base

c.

OH-(aq) is a strong base

d.

C2O(aq) is a weak base

e.

PO(aq) is a weak base

 

The solubilities of the above (b – e) will increase in acidic solution.  Only (a), which contains an extremely weak base (I- is the conjugate base of the strong acid HI) is unaffected by the acid solution.

17.70

In water:

                                                                Mg(OH)2  Mg2+  +  2OH-

                                                                                              s                2s

 

Ksp  =  4s3  =  1.2 ´ 10-11

 

s  =  1.4 ´ 10-4 M

 

In a buffer at pH  = 9.0

 

[H+]  =  1.0 ´ 10-9

 

[OH-]  =  1.0 ´ 10-5

 

1.2 ´ 10-11  =  (s)(1.0 ´ 10-5)2

 

s  =  0.12 M

17.71

From Table 17.4, the value of Ksp for iron(II) hydroxide is 1.6 ´ 10-14.

a.

At pH = 8.00, pOH = 14.00 - 8.00 = 6.00, and [OH-] = 1.0 ´ 10-6 M

 

 

The molar solubility of iron(II) hydroxide at pH = 8.00 is 0.016 M or 1.6 ´ 10-2 M

b.

At pH = 10.00, pOH = 14.00 - 10.00 = 4.00, and [OH-] = 1.0 ´ 10-4 M

 

 

The molar solubility of iron(II) hydroxide at pH = 10.00 is 1.6 ´ 10-6 M.

17.72

The solubility product expression for magnesium hydroxide is

 

Ksp  =  [Mg2+][OH-]2  =  1.2 ´ 10-11

 

We find the hydroxide ion concentration when [Mg2+] is 1.0 ´ 10-10 M

 

 

Therefore the concentration of OH- must be slightly greater than 0.35 M.

17.73

We first determine the effect of the added ammonia.  Let's calculate the concentration of NH3.  This is a dilution problem.

 

MiVi  =  MfVf

 

(0.60 M)(2.00 mL)  =  Mf(1002 mL)

 

Mf  =  0.0012 M NH3

 

Ammonia is a weak base (Kb  =  1.8 ´ 10-5).

 

                                       

Initial (M):

0.0012

 

 

0

0

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.0012 - x

 

 

x

x

 

 

 

 

Solving the resulting quadratic equation gives x = 0.00014, or [OH-] = 0.00014 M

 

This is a solution of iron(II) sulfate, which contains Fe2+ ions.  These Fe2+ ions could combine with OH- to precipitate Fe(OH)2.  Therefore, we must use Ksp for iron(II) hydroxide.  We compute the value of Qc for this solution.

 

Fe(OH)2(s)  Fe2+(aq) + 2OH-(aq)

 

Q  =  [Fe2+]0[OH-]  =  (1.0 ´ 10-3)(0.00014)2  =  2.0 ´ 10-11

 

Note that when adding 2.00 mL of NH3 to 1.0 L of FeSO4, the concentration of FeSO4 will decrease slightly.  However, rounding off to 2 significant figures, the concentration of 1.0 × 10-3 M does not change.  Q is larger than Ksp [Fe(OH)2] = 1.6 ´ 10-14.  The concentrations of the ions in solution are greater than the equilibrium concentrations; the solution is saturated.  The system will shift left to reestablish equilibrium; therefore, a precipitate of Fe(OH)2 will form.

17.74

First find the molarity of the copper(II) ion

 

 

 

Because of the magnitude of the Kf for formation of Cu(NH3), the position of equilibrium will be far to the right.  We assume essentially all the copper ion is complexed with NH3.  The NH3 consumed is             4 ´ 0.0174 M = 0.0696 M.  The uncombined NH3 remaining is (0.30 - 0.0696) M, or 0.23 M.  The equilibrium concentrations of Cu(NH3) and NH3 are therefore 0.0174 M and 0.23 M, respectively.  We find [Cu2+] from the formation constant expression.

 

 

[Cu2+]  =  1.2 ´ 10-13 M

17.75

Strategy:

The addition of Cd(NO3)2 to the NaCN solution results in complex ion formation.  In solution, Cd2+ ions will complex with CN- ions.  The concentration of Cd2+ will be determined by the following equilibrium

 

Cd2+(aq) + 4CN-(aq)   Cd(CN)

 

From Table 17.5 of the text, we see that the formation constant (Kf) for this reaction is very large (Kf  =  7.1 ´ 1016).  Because Kf is so large, the reaction lies mostly to the right.  At equilibrium, the concentration of Cd2+ will be very small.  As a good approximation, we can assume that essentially all the dissolved Cd2+ ions end up as Cd(CN) ions.  What is the initial concentration of Cd2+ ions?  A very small amount of Cd2+ will be present at equilibrium.  Set up the Kf expression for the above equilibrium to solve for [Cd2+].

Solution:

Calculate the initial concentration of Cd2+ ions.

 

 

If we assume that the above equilibrium goes to completion, we can write

 

 

                                       

Initial (M):

4.2 ´ 10-3

0.50

 

0

Change (M):

-4.2 ´ 10-3

-4(4.2 ´ 10-3)

 

+4.2 ´ 10-3

Equilibrium (M):

0

0.48

 

4.2 ´ 10-3

 

 

To find the concentration of free Cd2+ at equilibrium, use the formation constant expression.

 

 

Rearranging,

 

 

Substitute the equilibrium concentrations calculated above into the formation constant expression to calculate the equilibrium concentration of Cd2+.

 

 

[Cd(CN)] =  (4.2 ´ 10-3 M) - (1.1 ´ 10-18)  =  4.2 ´ 10-3 M

 

[CN-]  =  0.48 M + 4(1.1 ´ 10-18 M)  =  0.48 M

Think About It:

Substitute the equilibrium concentrations calculated into the formation constant expression to calculate Kf.  Also, the small value of [Cd2+] at equilibrium, compared to its initial concentration of 4.2 ´ 10-3 M, certainly justifies our approximation that almost all the Cd2+ ions react.

17.76

The reaction

 

Al(OH)3(s) + OH-(aq)   Al(OH)(aq)

 

is the sum of the two known reactions

 

                                Al(OH)3(s)   Al3+(aq) + 3OH-(aq)                                Ksp  =  1.8 ´ 10-33

 

                                Al3+(aq) + 4OH-(aq)   Al(OH)(aq)                           Kf  =  2.0 ´ 1033

 

The equilibrium constant is

 

 

When pH = 14.00, [OH-] = 1.0 M, therefore

 

[Al(OH)]  =  K[OH-]  =  (3.6)(1 M)  =  3.6 M

 

This represents the maximum possible concentration of the complex ion at pH 14.00.  Since this is much larger than the initial 0.010 M, the complex ion will be the predominant species.

17.77

Silver iodide is only slightly soluble.  It dissociates to form a small amount of Ag+ and I- ions.  The Ag+ ions then complex with NH3 in solution to form the complex ion Ag(NH3).  The balanced equations are:

 

AgI(s)   Ag+(aq) + I-(aq)                                               Ksp  =  [Ag+][I-]  =  8.3 ´ 10-17

Ag+(aq) + 2NH3(aq)   Ag(NH3)(aq)                      

Overall:  AgI(s) + 2NH3(aq)   Ag(NH3)(aq) + I-(aq)             K  =  Ksp ´ Kf  =  1.2 ´ 10-9

 


If s is the molar solubility of AgI then,

 

                                       

Initial (M):

 

1.0

 

0.0

0.0

Change (M):

-s

-2s

 

+s

+s

Equilibrium (M):

 

(1.0 - 2s)

 

s

s

 

 

 

 

 

 

 

 

Because Kf is large, we can assume all of the silver ions exist as Ag(NH3).  Thus,

 

                                                                          [Ag(NH3)]  =  [I-]  =  s

 

We can write the equilibrium constant expression for the above reaction, then solve for s.

 

 

s  =  3.5 ´ 10-5 M

 

At equilibrium, 3.5 ´ 10-5 moles of AgI dissolves in 1 L of 1.0 M NH3 solution.

17.78

Strategy:

The formula of hydroxyapatite is given in Section 17.4:  Ca5(PO4)3OH.  Write the dissociation equation for Ca5(PO4)3OH and solve for molar solubility using the equilibrium expression. 

Setup:

The equation for the dissociation of Ca5(PO4)3OH is

 

Ca5(PO4)3OH(s)   5Ca2+(aq) + 3PO(aq) + OH–(aq)

 

and the equilibrium expressions is Ksp = [Ca2+]5[PO]3[OH–]. 

Solution:

 

                                       

Initial (M):

 

 

0

0

0

Change (M):

 

 

+5s

+3s

+s

Equilibrium (M):

 

 

5s

3s

s

 

Therefore,

2 × 10–59 = (5s)5(3s)3(s)

 

2 × 10–59  = (3125s5)(27s3)(s)

 

2.4 × 10–64  = s9

 

s = 9 × 10–8  M

17.79

Strategy:

Substitute the molar solubility into the equilibrium expression to determine Ksp.

Setup:

s = 7 × 10–8  M

 

The equation and the equilibrium expression for the dissociation of Ca5(PO4)3F are:

 

Ca5(PO4)3F(s)   5Ca2+(aq) + 3PO