Homework Problems

Chapter 16 Homework Problems: 2, 4, 16, 18, 24, 27, 30a, 36, 39a,b, 46, 54, 64, 67, 78, 84, 86, 90, 96, 97 (but for Cu(OH) 2 ), 106

CHAPTER 16

Aqueous Ionic Equilibrium

Neutralization Reaction

A neutralization reaction is the reaction of an acid with a base, usually giving salt plus water as products.

For three types of neutralization reactions the reaction goes essentially to completion: strong acid + strong base HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(  ) strong acid + weak base HCl(aq) + NH 3 (aq)  NH 4 Cl(aq) weak acid + strong base HF(aq) + NaOH(aq)  NaF(aq) + H 2 O(  )

Buffers

A buffer solution is a solution whose pH does not change significantly upon the addition of a small amount of a strong acid or strong base.

Buffer solutions contain a substantial amount of a weak acid and a weak base, usually a weak acid-weak base conjugate pair.

Buffer solutions are important for systems where a particular value for pH is required, such as in biological systems.

How Buffers Work

Consider a buffer solution containing a weak acid “HA” and the conjugate base of the weak acid “A ”. What will happen if we add a small amount of a strong acid (HCl) or strong base (NaOH)?

strong acid HCl(aq) + A (aq)  HA(aq) + Cl (aq) strong base NaOH(aq) + HA(aq)  Na + (aq) + A (aq) + H 2 O(  ) Buffer solution converts the strong acid or base into a weak acid or base. By doing so, they minimize the effect of these additions on pH.

Example of Buffering

Consider 1.000 L of each of the following two solutions.

Solution A. 0.100 M CH 3 COOH and 0.200 M CH 3 COONa.

Solution B. Pure water Find the initial pH and the pH after addition of 0.0010 moles HCl to each of the above solutions.

Solution A. 0.100 M CH 3 COOH and 0.200 M CH 3 COONa.

CH CH 3 3 COONa(aq)  COOH(aq) + H 2 Na + (aq) + CH 3 COO (aq) O(  )  H 3 O + (aq) + CH 3 COO (aq) K a = [H 3 O + ] [CH 3 COO ] = 1.8 x 10 -5 [CH 3 COOH] Substance Initial H 3 O + CH 3 COO 0.000

0.200

CH 3 COOH 0.100

Change x x - x Equilibrium x 0.200 + x 0.100 - x

K a = [H 3 O + ] [CH 3 COO ] = (x) (0.200 + x) = 1.8 x 10 -5 [CH 3 COOH] (0.100 - x) Assume x << 0.100

(x) (0.200) = 1.8 x 10 -5 (0.100) x = (1.8 x 10 -5 ) (0.100) = 9.0 x 10 -6 (0.200) pH = - log 10 (9.0 x 10 -6 ) = 5.05

Is 9.0 x 10 -6 << 0.100? Yes (at least 10 times smaller).

Now, consider the effect of adding 0.0010 moles of HCl (equi valent to 0.0010 M HCl, since 1.00 L solution).

initial after rxn.

HCl(aq) + CH 3 COO (aq) 0.0010 M 0.2000 M 0.0000 M 0.1990 M  CH 3 COOH(aq) + Cl (aq) 0.1000 M 0.1010 M 0.0010 M CH 3 COOH(aq) + H 2 O(  )  H 3 O + (aq) + CH 3 COO (aq) K a = [H 3 O + ] [CH 3 COO ] = 1.8 x 10 -5 [CH 3 COOH] Substance Initial After HCl H 3 O + CH 3 COO 0.000

0.200

CH 3 COOH 0.100

0.000

0.199

0.101

Change x x - x Equilibrium x 0.199 + x 0.101 – x

K a = (x) (0.199 + x) = 1.8 x 10 -5 (0.101 – x) Assume x << 0.100, then (x) (0.199) = 1.8 x 10 -5 (0.101) x = (0.101) (1.8 x 10 -5 ) = 9.1 x 10 -6 (0.199) pH = - log 10 (9.1 x 10 -6 ) = 5.04

The pH decreased from 5.05 to 5.04, a change of 0.01 pH unit.

Solution B (pure water) pure water. pH = 7.00

After addition of 0.0010 moles of HCl we simply have a 0.001 M solution of HCl, a strong acid.

So pH = - log 10 (0.0010) = 3.00

We may summarize the results as follows: solution A solution B initial pH pH after addn. of 5.05

5.04

7.00

3.00

0.0010 mol HCl The pH of solution A (the buffer solution) changed by only a small amount after addition of the HCl.

The pH of solution B (unbuffered water) changed by 4 pH units (a change in H 3 O + concentration of a factor of 10,000).

Henderson Equation

The Henderson (or Henderson-Hasselbalch) equation applies for solutions of a weak acid/conjugate base pair.

We may develop the equation as follows.

Consider a weak acid HA HA(aq) + H 2 O(  ) K a = [H 3 O + ] [A ] [HA]  so H 3 O + (aq) + A (aq) [H 3 O + ] = K a [HA] [A ] Therefore - log 10 [H 3 O + ] = - log 10 K a - log 10 ([HA]/[A ]) pH = pK a - log 10 ([HA]/[A ]) , or pH = pK a + log 10 ([A ]/[HA]) , the Henderson equation.

Properties of the Henderson Equation

1) The Henderson equation applies for solutions of a weak acid/conjugate base pair.

2) There must be appreciable amounts (at least 0.001 M) of both the weak acid and the conjugate base present in the system.

3) If the weak acid and conjugate base concentrations are the same ([HA] = [A ]), then pH = pK a + log 10 ([A ]/[HA]) = pK a + log 10 (1) = pK a This suggests that the best buffers are those where pH ~ pK a . In fact, buffers work best if the pK a of the weak acid is within about 1 pH unit of the desired buffer pH.

4) While buffers are often prepared by combining a weak acid and conjugate base pair, they can also be prepared by adding a small amount of strong base to a weak acid solution, or a small amount of strong acid to a weak base solution.

Example: We have 0.500 M stock solutions of the following: CH 3 COOH NaCH 3 COO K a = 1.8 x 10 -5 HF NaF K a = 3.5 x 10 -4 HOCl NaOCl K a = 3.5 x 10 -8 pK a = 4.74

pK a = 3.46

pK a = 7.46

How would you prepare 1.000 L of a pH = 3.75 buffer solution?

Example: We have 0.500 M stock solutions of the following: CH 3 COOH NaCH 3 COO K a = 1.8 x 10 -5 HF NaF K a = 3.5 x 10 -4 HOCl NaOCl K a = 3.5 x 10 -8 pK a = 4.74

pK a = 3.46

pK a = 7.46

How would you prepare 1.000 L of a pH = 3.75 buffer solution?

Which acid/conjugate base pair should we use?

We should use the one where the pK a is as close as possible to the pH of the buffer we wish to make. That would be the HF/NaF system.

Now, let x = mL of the HF stock solution (1000 - x) = mL of the NaF stock solution

Then, in the 1.000 L buffer solution (using M i V i = M f V f ) [HF] = (0.500 M) x [F ] = (0.500 M) (1000 - x) 1000 1000 pH = pK a + log 10 ([A ]/[HA]) 3.75 = 3.46 + log 10 (0.500)[(1000 - x)/1000] (0.500) (x/1000) 3.75 = 3.46 + log 10 [(1000 - x)/x] 3.75 - 3.46 = 0.29 = log 10 [(1000 - x)/x] 10 0.29

= 1.95 = (1000 - x)/x 1.95 x = 1000 - x 2.95 x = 1000 ; x = 1000/2.95 = 339.

So 339 mL of the HF solution and 661 mL of the NaF solution.

Titration

By titration, we mean the reaction of a fixed amount of one sub stance by the slow addition of a second substance.

In an acid-base titration we slowly add an acid to a fixed amount of base, or we slowly add a base to a fixed amount of acid.

We use titrations to determine the concentration of acid or base in a stock solution. We may also be interested in the titration curve, a plot of pH vs mL of added titrant (solution being slowly added).

Equivalence point.

The equi valence point in a titration is the point where sufficient titrant has been added to have a complete reaction with no excess acid or base.

Indicator

An indicator is a substance (usually a weak acid) that changes colors at a particular pH.

HInd(aq) + H 2 O(  ) color 1  H 3 O + (aq) + Ind (aq) color 2 We often use indicators in titrations. We choose an indicator so that the end point (point where color change occurs) is as close as possible to the equivalence point for the titration. Based on the Henderson equation pH = pK a + log 10 ([A ]/[HA]) we expect the color change will occur at pH  pK Ind .

Table 16.1, page 747 of Tro.

In choosing an indicator we would like the color change to be as close as possible to the equivalence point for the titration.

Titration of a Strong Base With a Strong Acid

Consider the titration of 25.00 mL of a 0.0100 M solution of NaOH (a strong base) with a 0.0100 M solution of HCl (a strong acid).

The reactions that will occur are as follows: NaOH(aq) + HCl(aq) NaOH(aq)  NaCl(aq) + H 2 O(  )  Na + (aq) + OH (aq) HCl(aq)  H + (aq) + Cl (aq) neutralization if excess base if excess acid If there is excess NaOH the pH will be greater than 7.0. If there is excess HCl the pH will be less than 7.0. At the equivalence point the moles of NaOH and HCl will be equal, and the solution will be neutral (pH = 7.0).

The equivalence point occurs at pH = 7.0.

A good indicator would be one where the color change occurs near pH = 7.0 (such as bromthymol blue).

Notice that the titration curve is steep (nearly vertical) near the equivalence point.

Note that we can calculate the pH at every point in the titration (which was done in generating the above titration curve).

Titration of a Strong Acid With a Strong Base

Consider the same system, but the case where 25.00 mL of a 0.0100 M solution of HCl is titrated with a 0.0100 M solution of NaOH.

The reactions taking place are the same, but now the solution is initially acidic.

Titration of a Weak Acid With a Strong Base

CH 3 Consider the titration of 25.00 mL of a 0.0100 M solution of COOH (a weak acid; K NaOH (a strong base).

a = 1.8 x 10 -5 ) with a 0.0100 M solution of The reactions that will occur are as follows: CH 3 COOH(aq) + NaOH(aq)  H 2 O(  ) + Na + (aq) + CH 3 COO (aq) neutralization CH 3 COOH(aq) + H 2 O(  )  H 3 O + (aq) + CH 3 COO (aq) if excess weak acid NaOH(aq)  Na + (aq) + OH (aq) if excess base

CH 3 COOH(aq) + NaOH(aq)  H 2 O(  ) + Na + (aq) + CH 3 COO (aq) neutralization CH 3 COOH(aq) + H 2 O(  )  H 3 O + (aq) + CH 3 COO (aq) if excess weak acid NaOH(aq)  Na + (aq) + OH (aq) if excess base If there is excess CH 3 COOH then there will usually be a buffer solution with both CH 3 COOH and CH 3 COO present. If there is excess base then the pH will be controlled by the concentration of excess strong base.

Note that at the half-equivalence point pH  pK a , since pH = pK a + log 10 ([A ]/[HA])

At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) pH  pK a for the weak acid. Notice that there will be a region where the solution forms a buffer solution. Note that the pH at the equivalence point is greater than 7.0.

Indicator - phenolphthalein

Titration of a Weak Base With a Strong Acid

Consider the titration of 25.00 mL of a 0.0100 M solution of NH 3 (a weak base; K b = 1.8 x 10 -5 ) with a 0.0100 M solution of HCl (a strong acid).

The reactions that will occur are as follows: NH 3 (aq) + HCl(aq)  NH 3 (aq) + H 2 O(  )  NH 4 + (aq) + Cl (aq) neutralization NH 4 + (aq) + OH (aq) if excess weak base HCl(aq)  H + (aq) + Cl (aq) if excess acid If there is excess NH 3 with both NH 3 and NH 4 + then there will usually be a buffer solution present. If there is excess acid then the pH will be controlled by the concentration of excess strong acid.

As was the case for the titration of a weak acid with a strong base, at the half equivalence point pH  pK a (this will be for NH 4 + , the conjugate acid of NH 3 ).

At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) pH  pK a for the conjugate acid of the weak base. Notice that there will be a region where the solution forms a buffer solution. Note that the pH at the equivalence point is less than 7.0.

Indicator - chlorophenol red

Titration of a Polyprotic Acid With a Strong Base

Consider the titration of 1.000 L of a 1.000 M solution of a diprotic acid (H 2 A) with NaOH, a strong base.

Solubility Product

We previously divided ionic compounds into two general categories: insoluble - does not dissolve in water soluble - dissolves in water However, “insoluble” ionic compounds usually dissolve in water to a very small extent.

The solubility product, K sp , is defined as the equilibrium constant for the solubility reaction.

Examples: AgCl AgCl(s)  Ag + (aq) + Cl (aq) CaF 2 CaF 2 (s)  Ca 2+ (aq) + 2 F (aq) K K sp sp = [Ag = [Ca + 2+ ] [Cl ] [F ] ] 2 Since our reactant is a solid, equilibrium requires only that there be some solid present in the system.

Solubility and Molar Solubility

There are two terms that are related to the solubility product.

Solubility (solubility by mass) - The number of grams of a compound that will dissolve per 1.00 L of water.

Molar solubility - The number of moles of a compound that will dissolve per 1.00 L of water.

Of course, our previous methods for indicating solubility (such as grams per 100 grams solvent) are still valid.

Example: The solubility product for lead II chloride (PbCl 2 , MW = 278.1 g/mol) is K sp = 1.2 x 10 -5 at T = 25.

 C. What are the solubility by mass and the molar solubility of lead II chloride?

Example: The solubility product for lead II chloride (PbCl 2 , MW = 278.1 g/mol) is K sp = 1.2 x 10 -5 at T = 25.

 C. What are the solubility by mass and the molar solubility of lead II chloride?

PbCl 2 (s)  Pb 2+ (aq) + 2 Cl (aq) K sp = [Pb 2+ ] [Cl ] 2 = 1.2 x 10 -5 Substance Initial Change Equilibrium Pb 2+ Cl 0.00

0.00

(x) (2x) 2 = 4x 3 = 1.2 x 10 -5 + x + 2x x 3 = 3.0 x 10 -6 ; x = (3.0 x 10 -6 ) 1/3 = 1.44 x 10 -2 molar solubility = 1.44 x 10 -2 moles/L solubility = 1.44 x 10 -2 mol 278.1 g = 4.01 g x 2x 1 L 1 mol L

Factors Affecting Solubility

1) Temperature. By Le Chatlier’s principle, we may make the following general statements If the solubility reaction is exothermic (  H < 0) then solubility decreases as temperature increases.

AB(s)  A + (aq) + B (aq) + “heat” If the solubility reaction is endothermic (  H > 0) then solubility increases as temperature increases.

AB(s) + “heat”  A + (aq) + B (aq) 2) pH.

containing OH This will be true for hydroxide compounds (those ion) or compounds containing weak base anions (F CO 3 2 , S 2 , PO 3 3 , and other conjugate bases of weak acids).

, CN , Example: What is the maximum concentration of Ni 2+ ion in a solution with pH = 10.00? K sp (Ni(OH) 2 ) = 5.5 x 10 -16 .

Example: What is the maximum concentration of Ni 2+ solution with pH = 10.00? K sp (Ni(OH) 2 ) = 5.5 x 10 -16 .

ion in a Ni(OH) 2  Ni 2+ (aq) + 2 OH (aq) K sp = [Ni 2+ ] [OH ] 2 = 5.5 x 10 -16 Since pH = 10.00, pOH = 14.00 - 10.00 = 4.00.

So [OH ] = 10 -4.00

= 1.0 x 10 -4 M K sp = [Ni 2+ ] [OH ] 2 [Ni 2+ ] = K sp = 5.5 x 10 -16 = 5.5 x 10 -8 M [OH ] 2 (1.0 x 10 -4 ) 2 Note that it is possible for [Ni 2+ ] to be  5.5 x 10 -8 M.

3) Common ion effect. The common ion effect refers to the fact that in equilibrium reactions all that matters are the concentrations of the ions involved in the equilibrium. The source of the ions does not matter.

If a common ion is already present in solution this can affect the solubility of a compound.

Example: What is the molar solubility of AgCl in a) pure water and b) in a 0.100 M solution of NaCl. K sp (AgCl) = 1.8 x 10 -10 .

Example: What is the molar solubility of AgCl in a) pure water and b) in a 0.100 M solution of NaCl. K sp (AgCl) = 1.8 x 10 -10 .

AgCl(s)  Ag + (aq) + Cl (aq) K sp = [Ag + ] [Cl ] = 1.8 x 10 -10 a) pure water compound Ag + Cl initial 0.00

0.00

(x) (x) = x 2 = 1.8 x 10 -10 change + x + x x = (1.8 x 10 -10 ) 1/2 = 1.34 x 10 -5 molar solubility = 1.34 x 10 -5 mol/L.

equilibrium x x

b) 0.100 M solution of NaCl NaCl(s)  Na + (aq) + Cl (aq) compound Ag + initial 0.00

change + x equilibrium x Cl 0.100

+ x 0.100 + x (x) (0.100 + x) = 1.8 x 10 -10 Assume x << 0.100

(x) (0.100) = 1.8 x 10 -10 x= (1.8 x 10 -10 ) = 1.8 x 10 -9 (0.100) molar solubility = 1.8 x 10 -9 mol/L.

This is much smaller than the molar solubility we found for silver chloride in pure water (1.34 x 10 -5 mol/L).

Solubility and the Reaction Quotient

Consider a solution where only ions are present (no solid initially present). As a specific example consider a 1:1 ionic compound MX MX(s)  M + (aq) + X (aq) Q sp = [M + ][X ] There are three possibilities for Q sp , the reaction quotient If Q sp dissolve).

< K sp the solution is unsaturated (more solid could If Q sp dissolve).

= K sp the solution is saturated (no more solid could If Q sp > K sp the solution is supersaturated (a precipitate will form as the system goes to equilibrium).

Selective Precipitation

We may use differences in solubility to remove specific cations from a mixture by selective precipitation, the formation of a precipitate with one ion in the mixture.

Example: Consider a solution containing 0.0100 M Mg 2+ 0.0100 M Cu 2+ ion. Can we separate these ions?

ion and

Example: Consider a solution containing 0.0100 M Mg 2+ 0.0100 M Cu 2+ ion. Can we separate these ions?

ion and Consider adding OH ion.

Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH (aq) Cu(OH) K sp = [Mg 2+ ] [OH ] 2 = 5.6 x 10 -12 2 (s)  Cu 2+ (aq) + 2 OH (aq) K sp = [Cu 2+ ] [OH ] 2 = 1.6 x 10 -19 Mg Cu [OH ] = {K sp /[Mg 2+ ]} 1/2 = {5.6 x 10 -12 /0.01} 1/2 = 2.4 x 10 -5 M [OH ] = {K sp /[Cu 2+ ]} 1/2 = {1.6 x 10 -19 /0.01} 1/2 = 4.0 x 10 -9 M So if OH is added to the solution Cu(OH) 2 precipitate when [OH ] reaches 4.0 x 10 -9 . No Mg(OH) 2 until [OH ] reaches 2.4 x 10 -5 will begin to will precipitate M. So we can separate most of the Cu 2+ ion out of solution.

Complex Ion Formation

Some metal cations will react with small molecules or ions (NH 3 , CN , OH ) to form complex ions. A complex ion is defined as a metal cation bonded to one or more small molecules or ions.

For example Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation of a complex ion is called the formation constant (or stability constant), K f K f = [Ag(NH 3 ) 2 + ] [Ag + ] [NH 3 ] 2 Formation constants are usually large numbers (meaning complex ion formation is favored when the reactants are present).

End of Chapter 16

Tony Cortino: Where did you learn to dance like that?

Pepper: Julliard. I wanted to be a research chemist, but my legs were too long.

- dialogue from the movie Mafia!