CHM 1046. Chapter 16 Homework Solutions

CHM 1046. Chapter 16 Homework Solutions. Problems: 2, 4, 16, 18, 24, 27, 30a, 36, 39a,b, 46, 54, 64, 67, 78, 84, 86, 90, 96, 97 (but with Cu(OH)2), 106

2) A buffer is a solution that is resistant to changes in pH. Buffers work by converting strong acids or bases into weak acids or bases.

Consider a buffer that contains a weak acid (HA) and its conjugate base (A -). If we add a strong acid like HCl, or a strong base like NaOH, the reactions that occur are as follows

HCl(aq) + A-(aq)  HA(aq) + Cl-(aq)

- NaOH(aq) + HA  H2O(ℓ) + A (aq)

4) The Henderson-Hasselbach equation is

pH = pKa + log10{[base]/[acid]}

It is useful in preparing a buffer solution with a particular value for pH.

16) This is because for each neutralization it takes the same amount of base to completely react with the particular proton being donated by the acid.

18) An indicator is a substance that changes color depending on pH. If the color change for the indicator (marking the end point for the titration) takes place near the value of mL added titrant where the equivalence point occurs, the color change can be used to determine the equivalence point.

24) Selective precipitation refers to the process where an anion is added to a solution that causes one cation to precipitate while other cations remain in solution.

27) The ionization reaction for HNO2 is

+ - HNO2(aq) + H2O(ℓ)  H3O (aq) + NO2 (aq)

Le Chatlier's principle can be used to decide whether the above equilibrium will be shifted left or right

a) Added NaCl has no effect.

b) Added KNO3 has no effect.

- + c) Added NaOH increases OH concentration, and so decreases H3O concentration. The reaction will respond by shifting from left to right (and so more ionization).

1 - d) Added NaNO2 increases NO2 concentration. The reaction will respond by shifting from right to left (and so less ionization).

30) a) Reaction is

+ - HC2H3O2(aq) + H2O(ℓ)  H3O (aq) + C2H3O2 (aq)

+ - -5 Ka = [H3O ] [C2H3O2 ] = 1.8 x 10 [HC2H3O2]

Initial Change Equilibrium

HC2H3O2 0.175 - x 0.175 - x + H3O 0.0 x x - C2H3O2 0.110 x 0.110 + x

(x) (0.110 + x) = 1.8 x 10-5 (0.175 - x)

Assume x << 0.110. Then

x (0.110) = 1.8 x 10-5 ; so x = (0.175) (1.8 x 10-5) = 2.86 x 10-5 (0.175) (0.110)

-5 pH = - log10(2.86 x 10 ) = 4.54

Note that one could get the same result more quickly by using the Henderson equation

-5 pH = pKa + log10{[base]/[acid]} = - log10(1.8 x 10 ) + log10(0.110/0.175) = 4.54

+ 36) Ammonia (NH3) is a weak base, and ammonium ion (NH4 ) is a weak acid. Consider what happens when a strong acid (HCl) or a strong base (NaOH) is added to a solution containing the above substances.

+ - HCl(aq) + NH3(aq)  NH4 (aq) + Cl (aq) effect of added strong acid

+ NaOH(aq) + NH4 (aq)  H2O(ℓ) + NH3(aq) effect of added strong base

2 39) The Henderson equation is

-8 - a) Ka(HClO) = 2.9 x 10 [HClO] = 0.125 M [ClO ] = 0.150 M

-8 pH = - log10(2.9 x 10 ) + log10{(0.150)/(0.125)} = 7.62

-4 + b) Kb(C2H5NH2) = 5.6 x 10 ; so Ka for C2H5NH3 can be found by

-14 + -14 -11 Ka Kb = 1.0 x 10 , so Ka(C2H5NH3 ) = 1.0 x 10 = 1.79 x 10 5.6 x 10-4

+ [C2H5NH2] = 0.175 M [C2H5NH3 ] = 0.150 M

-11 pH = - log10(1.79 x 10 ) + log10{(0.175)/(0.150)} = 10.82

46) The Henderson equation is

-5 + Kb(NH3) = 1.76 x 10 ; so Ka for NH4 can be found by

-14 + -14 -10 Ka Kb = 1.0 x 10 , so Ka(C2H5NH3 ) = 1.0 x 10 = 5.68 x 10 1.76 x 10-5

-10 9.55 = - log10(5.68 x 10 ) + log10{[base]/[acid]} = 9.25 + log10{[base]/[acid]}

So log10{[base]/[acid]} = 9.55 - 9.25 = 0.30 Take the inverse log10, to get

[base]/[acid] = 100.30 = 2.00

And so [acid] = [base]/2.00 = 0.155 M/2.00 = 0.0775 M

M(NH4Cl) = 53.49 g.mol

So g(NH4Cl) = 2.55 L soln 0.0775 mol 53.49 g = 10.57 g NH4Cl 1 L 1 mol

54) A buffer requires substantial amounts of a weak acid and its conjugate base (or a weak base and its conjugate acid).

a) Yes. You have HF (weak acid) and F-, from NaF (weak base).

b) No. HCl is a strong acid and so will not form a buffer.

3 c) Yes. You have more HF than NaOH. Since HF reacts with NaOH by the process

HF(aq) + NaOH(aq)  NaF(aq) + H2O(ℓ)

So the NaOH converts some of the HF (weak acid) into F- (weak base). Therefore you get a buffer.

+ d) Yes. You have CH3NH2 (weak base) and CH3NH3 (conjugate acid).

e) Yes. You have more CH3NH2 than HCl. Since CH3NH2 reacts with HCl by the process

CH3NH2(aq) + HCl(aq)  CH3NH3Cl(aq)

+ So the HCl converts some of the CH3NH2 (weak base) into CH3NH3 (weak acid). Therefore you get a buffer.

64) For a, pH  7.0 at the equivalence point

For b, pH  5.0 at the equivalence point

The titrant is a strong acid. Since a has a pH of about 7.0 at the equivalence point, the base in a must be a strong base (strong acid + strong base gives a pH of about 7.0 at equivalence). So b must be a weak base (strong acid + weak base gives a pH less than 7.0 at equialence).

67) a) Since HBr is a strong acid

+ Initial [H3O ] = Initial [HBr] = 0.175 M, so pH = - log10(0.175) = 0.76

b) The neutralization reaction is

HBr(aq) + KOH(aq)  KBr(aq) + H2O(ℓ)

Since this is a 1:1 acid:base reaction, it follows that at the equivalence point moles acid = moles base

(35.0 ml) (0.175 M) = (x mL) (0.200 M)

x = 0.175 M (35.0 mL) = 30.62 mL 0.200 M c) After 10.0 mL added base

Total volume = 35.0 mL + 10.0 mL = 45.0 mL

4 Initial moles acid = (0.0350 L) (0.175 M) = 0.00612 mol acid

Initial moles base = (0.0100 L) (0.200 M) = 0.00200 mol base

Moles excess acid = 0.0612 mol - 0.0200 mol = 0.00412 mol

So the concentration of excess acid is 0.0412 mol = 0.0916 M 0.0450 L

Since the excess acid is a strong acid, pH = - log10(0.0916) = 1.04

d) At the equivalence point neutralization has occurred, and so what is present is a KBr solution, which has no acid-base properties. So the pH will be 7.0.

e) Volume = volume at equivalence + excess volume = 35.0 mL = 30.62 mL + 5.0 mL = 70.6 mL

mol excess base = (0.005 L) (0.200 M) = 0.00100 mol

Concentration of excess base = 0.00100 mol = 0.0142 M 0.0706 L

Since the base is strong, pOH = - log10(excess base) = - log10(0.0142) = 1.84

And so pH = 14.00 - pOH = 14.00 - 1.84 = 12.16

78) The equivalence point occurs after ~ 33. mL of KOH has been added.

Moles acid = Moles base = (0.033 L) (0.105 M) = 3.46 x 10-3 mol

M = g = 0.446 g = 129 g/mol mol 3.46 x 10-3 mol

The pH at the half-equivalence point is approximately equal to the pKa for the weak acid. So pKa  4.0

84) a) Weak base, so pH at equivalence point will be < 7.0. I would use methyl red

(pKa  5.4)

b) Strong base, so pH at equivalence point will be 7.0. I would use bromthymol blue (pKa  6.6), though there are several other good indicators that could be used.

c) Weak base, so the same as in a (methyl red).

5 2+ 2- 2+ 2- 86) a) CaCO3(s)  Ca (aq) + CO3 (aq) Ksp = [Ca ] [CO3 ]

2+ - 2+ - 2 b) PbCl2(s)  Pb (aq) + 2 Cl (aq) Ksp = [Pb ] [Cl ]

+ - + - c) AgI(s)  Ag (aq) + I (aq) Ksp = [Ag ] [I ]

2+ 2- 2+ 2- 90) a) BaCrO4(s)  Ba (aq) + CrO4 (aq) Ksp = [Ba ] [CrO4 ]

2+ -5 2- -5 [Ba ] = 1.08 x 10 M [CrO4 ] = 1.08 x 10 M

-5 -5 -10 Ksp = (1.08 x 10 ) (1.08 x 10 ) = 1.17 x 10

+ 2- + 2 2- b) Ag2SO3(s)  2 Ag (aq) + SO3 (aq) Ksp = [Ag ] [SO3 ]

+ -5 -5 2- -5 [Ag ] = 2 (1.55 x 10 M) = 3.10 x 10 M [SO3 ] = 1.55 x 10 M

-5 2 -5 -14 Ksp = (3.10 x 10 ) (1.55 x 10 ) = 1.49 x 10

2+ - 2+ - c) Pd(SCN)2(s)  Pd (aq) + 2 SCN (aq) Ksp = [Pd ] [SCN ]

[Pd2+] = 2.22 x 10-8 M [SCN-] = 2 (2.22 x 10-8 M) = 4.44 x 10-8 M

-8 -8 2 -23 Ksp = (2.22 x 10 ) (4.44 x 10 ) = 4.38 x 10

2+ 2- -36 96) CuS(s)  Cu2+(aq) + S2-(aq)Ksp = [Cu ] [S ] = 1.27 x 10

a) pure water

Cu2+ 0 x x

S2- 0 x x

(x) (x) = x2 = 1.27 x 10-36 ; x = (1.27 x 10-36)1/2 = 1.13 x 10-18

Molar solubility is 1.13 x 10-18 M

2+ b) 0.25 M CuCl2 , which gives 0.25 M Cu

Cu2+ 0.25 x 0.25 + x

S2- 0 x x

6 (0.25 + x) (x) = 1.27 x 10-36 Assume x << 0.25, then

(0.25) (x) = 1.27 x 10-36 x = (1.27 x 10-36)/(0.25) = 5.08 x 10-36

2- c) 0.20 M K2S , which gives 0.20 M S

Cu2+ 0 x x

S2- 0.20 x 0.20 + x

(x) (0.20 + x) = 1.27 x 10-36 Assume x << 0.20, then

(x) (0.20) = 1.27 x 10-36 x = (1.27 x 10-36)/(0.20) = 6.35 x 10-36

97) We will do the problem with copper II hydroxide, Cu(OH)2

2+ - 2+ - 2 -20 Cu(OH)2(s)  Cu (aq) + 2 OH (aq) Ksp = [Cu ] [OH ] = 2.2 x 10

So [Cu2+] = 2.2 x 10-20 [OH-]2

a) pH = 4.00 ; pOH = 14.00 - 4.00 = 10.00 ; [OH-] = 10-10.00 = 1.00 x 10-10

[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 M (1.0 x 10-10)2

b) pH = 7.00 ; pOH = 14.00 - 7.00 = 7.00 ; [OH-] = 10-7.00 = 1.00 x 10-7

[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 x 10-6 M (1.0 x 10-7)2

c) pH = 9.00 ; pOH = 14.00 - 9.00 = 5.00 ; [OH-] = 10-5.00 = 1.00 x 10-5

[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 x 10-10 M (1.0 x 10-5)2

7 2+ - 2+ - 2 -5 106) a) BaF2(s)  Ba (aq) + 2 F (aq) Ksp = [Ba ] [F ] = 2.45 x 10

2+ [Ba ] = [Ba(NO3)2] = 0.35 M

[F-]2 = 2.45 x 10-5 = 2.45 x 10-5 = 7.0 x 10-4 ; [F-] = (7.0 x 10-4)1/2 = 2.6 x 10-2 [Ba2+] 0.35

So minimum NaF concentration = 2.6 x 10-2 M

2+ 2- 2+ 2- -5 b) CaSO4(s)  Ca (aq) + SO4 (aq) Ksp = [Ca ] [SO4 ] = 7.10 x 10

2+ [Ca ] = [CaI2] = 0.085 M

2- -5 -5 -4 [SO4 ] = 7.10 x 10 = 7.10 x 10 = 8.35 x 10 [Ca2+] 0.085

-4 So minimum K2SO4 concentration = 8.35 x 10 M

+ - + - -10 c) AgCl(s)  Ag (aq) + Cl (aq) Ksp = [Ag ] [Cl ] = 1.77 x 10

+ [Ag ] = [AgNO3] = 0.0018 M

[Cl-] = 1.77 x 10-10 = 1.77 x 10-10 = 9.83 x 10-8 [Ag+] 0.0018

So minimum RbCl concentration = 9.83 x 10-8 M