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Exam II
Chemical Equlibria LeChatlier’s Principle
Acid Dissociation Base Hydrolysis Titration Curves
Solubility Product pH Controlled Solubility
Amphoterism
Equilibria and LeChatliers Principle 1) For each of the following sets of compounds write the equilibrium reaction that would occur when the compounds are mixed together. a) HNO3 and NaBenz b) NH4Cl and KOH c) NaCN and NaOH 2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after the system reaches equilibrium? H2 + Cl2 ---> 2 HCl K = 150 2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been added to the container originally? 3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium concentration of all ionic species. Ka = 6.76x10-4 4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5 5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M KOH? Ka = 6.46x10-5 6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65 mL of HCl? Kb = 1.8x10-5 7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3
- be in a solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15 8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5 Ac- + H2O <---> HAc + OH-
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction, 2 NH3 ↔ N2 + 3 H2 Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total pressure) be in the system? 10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic acid? Ka = 1.8x10-5 11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5). 12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M. 13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and 0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M 14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10. 15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAsp- 1.38x10-4 HAsp- ↔ H+ + Asp2- 1.51x10-10 0 mL 200 mL 400 mL 700 mL 800 mL 900 mL
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a new buffer of pH 4.5?
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the following volumes of 1.20 M NaOH?
H3His ↔ H+ + H2His- pKa1 = 2.40 H2His- ↔ H+ + HHis2- pKa2 = 6.04 HHis2- ↔ H+ + His3- pKa3 = 9.33 0 mL 30 mL 50 mL 75 mL 100 mL 125 mL 135 mL 150 mL 175 mL
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH = 8.0? Kb = 1.8x10-5 18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. 19) Another way of making the buffer from 2 above would be to titrate the acid with a base. What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. 20) How much 0.5 M NaOH must be added to 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67 Solubility Products 21) Are the following molecules acidic, basic or neutral in aqueous solution? NaF acid basic neutral can't tell Cr(NO3)3 acid basic neutral can't tell KCl acid basic neutral can't tell NH4CN acid basic neutral can't tell 22) Are the following compounds soluble or insoluble in water? NaIO3 Soluble Insoluble FeSO4 Soluble Insoluble Cr(OH)3 Soluble Insoluble ScCl3 Soluble Insoluble PbSO4 Soluble Insoluble Na2O Soluble Insoluble CaCl2 Soluble Insoluble PbSO4 Soluble Insoluble
23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5 24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6 25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8 26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8 27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10 28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium? Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20 29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl. Assume a liter of solution and calculate the concentration of NH3(aq). AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10 Ag(NH3)2
+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9 30) Copper(I) ions in aqueous solution react with NH3 according to, Cu+ + 2 NH3 ↔ Cu(NH3)2
+ Keq = 6.3x1010 Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium concentration of NH3 is 0.185 M. 31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-). PbS ↔ Pb2+ + S2- Ksp = 8x10-28
H2S ↔ 2 H+ + S2- Keq = 1x10-20 32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6. Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-). 33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH = 3? CdS ↔ Cd2+ + S2- Ksp = 8x10-27 H2S ↔ 2H+ + S2- Keq = 1x10-20 34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated solution of H2S = 0.10 M at pH = 3? Ag2S ↔ 2 Ag+ + S2- Ksp = 1.6x10-49 H2S ↔ 2H+ + S2- Keq = 1x10-20
35) How much of each precipitate will form and what will the concentration of lead be if 0.75 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M Mg(NO3)2 and 0.5 M Pb(NO3)2?
MgOx(s) ↔ Mg2+ + Ox2- Ksp = 8.6x10-5 PbOx(s) ↔ Pb2+ + Ox2- Ksp = 4.8x10-10
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 9? Ag2CO3 ↔ 2 Ag+ + CO3
2- Ksp = 8.1x10-12 H2CO3 ↔ 2 H+ + CO3
2- Ka = 2.02x10-17
Equilibria and LeChatliers Principle 1) For each of the following sets of compounds write the equilibrium reaction that would occur when the compounds are mixed together.
a) HNO3 and NaBenz HBenz ↔ H+ + Benz- b) NH4Cl and KOH NH4OH ↔ NH4
+ + OH- c) NaCN and NaOH CN- + H2O ↔ HCN + OH-
2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after the system reaches equilibrium? H2 + Cl2 ↔ 2 HCl K = 150 x x 4 atm – 2x
Since the equilibrium constant is large (150) we must shift the ingredients to the right. This means that 2 atm H2 and 2 atm Cl2 will produce 4 atm HCl. Using this value we can now solve the equilibrium
You do not have to assume that 2x is small. You can solve this exactly. Take the square root of both sides and then solve for x. So, x = 0.2808 atm so the pressure of HCl = 4 atm – 2(0.2808 atm) = 3.44 atm
2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been added to the container originally?
x = 0.1633 atm H2 and Cl2 left after equilibrium. To make 2 atm of HCl, 1 atm H2 and 1 atm Cl2 must have been consumed. Therefore there must have been 1 atm + 0.1633 = 1.1633 atm H2 and Cl2 to start.
3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium concentration of all ionic species. Ka = 6.76x10-4 Note: HF is not ionic but was included. 100g NaF/42 g/mol = 2.381 mol NaF and 70g KOH/56.1 g/mol = 1.248 mol KOH
F- + H2O ↔ HF + OH- 114
14
10822.2][1076.6
101
]52.9[
]99.4][[
][
]][[ −−
−
−
−
=⇒=−
+= xHFx
x
x
xHF
F
OHHF
[H+] = 1x10-14/[OH-] = 1x10-14/[4.99 M] = 2x10-15 M H+
Ion Moles Before Rxn Moles After Rxn Conc (250 mL total) Na+ 2.381 2.381 9.52 M F- 2.381 2.381 9.52 M K+ 1.248 1.248 4.99 M
OH- 1.248 1.248 4.99 M HF 0 From equilibrium 2.822x10-11 H+ 0 From equilibrium 2x10-15
150]24[
2
2
=−x
x
150]2[
2
2
=x
atm
4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5 (1.2M HAc)(0.100 L) = 0.12 mol HAc and (2M)(0.020L) = 0.040 mol NaOH HAc + NaOH ↔ H2O + NaAc 0.12 0.040 0 -0.04 -0.040 +0.040 0.08 0 0.040
HAc ↔ H+ + Ac- 0.08-x x 0.04+x
55 106.3][108.1]080.0[
]040.0][[
][
]][[ −−+−+
=⇒=−
+= xHxx
xH
HAc
AcH pH = 4.44
5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M KOH? Ka = 6.46x10-5 Benz- + H2O ↔ OH- + HBenz 0.25 0.25 0.10 +0.10 -0.10 -0.10 0.35-x 0.15+x 0+x If you ignore x, [OH-] = 0.15 M and [H+] = 1x10-14/0.15 = 6.66x10-14 so pH = 13.18 6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65 mL of HCl? Kb = 1.8x10-5 If the NH4OH is completely neutralized, then the only thing left in solution is NH4Cl (an acid). The only question is, what is the concentration of the NH4Cl? Total volume = 100 mL + 65 mL
(0.50 M)(0.100 L) = 0.050 mol NH4OH = 0.050 mol NH4Cl NH4
+ ↔ H+ + NH3 0.3030-x x x
+−−+
+
=⇒=−
= H Mx10297.1x5.55x10x0.3030
x
][NH
]][NH[H 5102
4
3 pH = 4.89
ClNH M 0.3030L 0.165
ClNH mol 0.0504
4 =
7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a
solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15 [Ca2+] [IO3
-]2 = x(2x)2 = 4x3 = 1.5x10-15 x = 7.21x10-6 8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5 Ac- + H2O ↔ HAc + OH-
−−−−
=⇒=−
= OH Mx1045.7x5.55x10x0.1
x
][Ac
][HAc][OH 6102
- pH = 8.87
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction, 2 NH3 ↔ N2 + 3 H2 Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total pressure) be in the system?
2 NH3 ↔ N2 + 3 H2 3 – 2x x 3x ignore the 2x on the bottom and solve. x = 0.1778 atm Total pressure = 3-2x+x+3x = 3+2x = 3+2(0.1778) = 3.3556 atm
10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic acid? Ka = 1.8x10-5
(1 M)(0.250 L) = 0.250 mol HAc Use the HH equation pH = 4.74 + log 0.20/0.25 pH = 4.64
11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5). A 0.10 M Ca2+ solution will require. [Ca2+] [OH-]2 = [0.10 M] [OH-]2 = 7.88x10-6 [OH-] = 0.00888 M OH- to precipitate 0.10 M NH4OH can only produce, this is not enough to ppt. Ca2+
32
4
2
3
103)23(
27
)23(
)3( −=−
=−
xx
x
x
xx
−−+−
=⇒=−
= OH M 0.001342x1.8x10x0.1
x
OH][NH
]][NH[OH 52
4
4
12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M. Ac- + H+ → HAc 0.25 0.15 pH = 4.57 -0.15 -0.15 +0.15 0.10 0 0.15 13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and 0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M NH4OH ↔ NH4
+ + OH- 0.30 0.50 0.25 Shift +0.25 -0.25 -0.25 0.55 0.25 0 pH = 9.60 → trickle 0.55-x 0.25+x 0+x 14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10.
(0.5 M)(0.030 L) = 0.015 mol NaOH and (0.75 M)(0.120 L) = 0.090 mol HCN HCN + OH- ↔ H2O + CN- 0.09 0.015 -0.015 -0.015 +0.015 0.075 0 0.015 [H+] = 2.4x10-9 [OH-] = 1x10-14/[2.4x10-9] = 4.17x10-6 M OH-
Ion Moles Before Rxn Moles After Rxn Conc (150 mL total) Na+ 0.015 0.015 0.10 M OH- 0.015 From equilibrium 4.17x10-6 HCN 0.090 0.075 0.05 M H+ 0 From equilibrium 2.4x10-9 CN- 0 0.015 0.05 M
55 x107.2][H1.8x10x][0.15
x]][0.10[H[HAc]
]][Ac[H −+−+−+
=⇒=−
+=
−−−+−
=⇒=−+= OH M 1096.3x1.8x10x0.55x)x(0.25
OH][NH]][NH[OH 55
4
4 x
10108.4]075.0[
]015.0][[][
]][[ −+−+
=−
+= xx
xH
HCN
CNH
15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAsp- 1.38x10-4 pKA1 = 3.86 HAsp- ↔ H+ + Asp2- 1.51x10-10 pKA2 = 9.82
0 mL pH = 1.69 200 mL pH = 3.86 400 mL pH = 6.84 700 mL 300 mL into 2nd Region (1.2 M)(0.300 L) = 0.36 mol OH- pH = 10.30
800 mL 510-
-14-2 1062.6
1.51x101x10
Kb and M 0.50L 0.960
mol 0.48][Asp −==== x
x = 5.75x10-3 M OH- pH = 11.76 900 mL (0.100L)(1.20M) = 0.12 mol OH-
−−
= OH M 0.1132L 1.060OH mol 0.12
pH = 13.05
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a new buffer of pH 4.5? First, calculate the composition of the existing buffer, [HAsp-] = 6.90 mol Now find out how much acid must be added to make new buffer (pH is going down so add acid). x = 0.8792 mol H+ = (1.2 M)(V) V = 732.7 mL of 1.2 M HCl
+− =⇒=−
H M 0.002035x1.38x10x3
x 42
M1V1 = M2V2 (3 M)(0.160 L) = (1.20 M)(V2) V2 = 400 mL Also, (3 M)(0.160 L) = 0.48 mol H2Asp
282.986.3
2pKpK
pH A2A1 +=+=
0.360.480.36
log82.9pH−
+=
52
1062.650.0
−=−
xx
x
A
BpKapH log+=
5.0
][log86.35
−
+= HAsp
x
x
+−+=
5.090.6
log86.35.4
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the following volumes of 1.20 M NaOH?
H3His <---> H+ + H2His- pKa1 = 2.40 H2His- <---> H+ + HHis2- pKa2 = 6.04 HHis2- <---> H+ + His3- pKa3 = 9.33
0 mL pH = 1.248 30 mL (1.2 M)(0.030 L) = 0.036 mol OH- pH = 2.58 50 mL pH = 4.22 75 mL pH = 6.04 100 mL pH = 7.69 125 mL pH = 9.33 135 mL 35 mL into 3rd Region (1.2 M)(0.035 L) = 0.042 mol OH- pH = 9.70
150 mL 510-
-14-3 1014.2
4.68x10
1x10 Kb and M 0.2667
L 0.225
mol 0.060][His −==== x
x = 2,39x10-3 M OH- pH = 11.38 175 mL (0.025 L)(1.20M) = 0.030 mol OH-
−−
= OH M 0.120L 0.250
OH mol 0.030 pH = 13.08
M1V1 = M2V2 (0.8 M)(0.075 L) = (1.20 M)(V2) V2 = 50 mL Also, (0.8 M)(0.075 L) = 0.060 mol H3His
+− =⇒=−
H M 0.0564x8x109.3x8.0
x 32
204.640.2
2pKpK
pH A2A1 +=+=
233.904.6
2pKpK
pH A3A2 +=+=
0.0360.0600.036
log40.2pH−
+=
0.0420.0600.042
log33.9pH−
+=
52
1014.22667.0
−=−
xx
x
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH = 8.0? Kb = 1.8x10-5 Using the base version of the HH equation;
Adding 18.19 moles to 250 mL of water will make a solution that is 72.76M, so divide top and bottom by 100. Add 0.1819 mol NH4Cl and 0.010 mole NH4OH to 250 mL of water to make a buffer of pH = 8
18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. (0.25 M HAc)(0.100 L) = 0.025 mol HAc 19) Another way of making the buffer from 18 above would be to titrate the acid with a base. What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. (0.40 M)(V) = 0.003849 mol NaOH V = 9.62 mL 20) How much 0.5 M NaOH must be added to 80 mL of 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67 The pH is nearest the 2nd pKa so the buffer is in Region II. You must add, (0.75 M)(0.080 L) = (0.50 M)(V) V = 120 mL so add 120 mL to get to the starting point. = (0.50 M)(V) V = 103.25 mL Total volume = 103.25 mL + 120 mL = 223.35 mL NaOH
01.0
1819.0
1
19.18
OH][NH
Cl][NH
OH][NH
Cl][NHlog74.46
4
4
4
4 ==⇒+=
NaAc mol 0.00455xHAc) mol (0.025
(x)log74.44 =⇒+=
NaOH mol 0.003849xx)- (0.025
(x)log74.44 =⇒+=
NaOH mol 0.05163xx)- (0.060
(x)log21.78 =⇒+=
Solubility Products 21) Are the following molecules acidic, basic or neutral in aqueous solution? NaF acid basic neutral can't tell Cr(NO3)3 acid basic neutral can't tell KCl acid basic neutral can't tell NH4CN acid basic neutral can't tell 22) Are the following compounds soluble or insoluble in water? NaIO3 Soluble Insoluble FeSO4 Soluble Insoluble Cr(OH)3 Soluble Insoluble ScCl3 Soluble Insoluble PbSO4 Soluble Insoluble Na2O Soluble Insoluble CaCl2 Soluble Insoluble Hg2SO4 Soluble Insoluble 23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5 [Pb2+] [Br -]2 = x(2x)2 = 4x3 = 4x10-5 x = 2.15x10-2 M 24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6 [Ca2+] [0.50 M]2 = 5.5x10-6 x = 2.20x10-5 25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8 [Ca2+]3 [PO4
3-]2 = (3x)3(2x)2 = 108x5 = 5.87x10-8 x = 1.40x10-2 M 26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8 [Ca2+]3 [PO4
3-]2 = (3x)3(0.50)2 = 0.75x3 = 5.87x10-8 x = 4.278x10-3 M 27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10 [Ag+] [1 M] = 1.8x10-10 [Ag+] = 1.8x10-10 28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium? Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20 [Cu2+] [OH-]2 = [0.02 M] [OH-]2 = 2.2x10-20 [OH-] = 1.05x10-9 pH = 6.02
29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl. Assume a liter of solution and calculate the concentration of NH3(aq). AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10 Ag(NH3)2
+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9 If you reverse the bottom reaction and add it to the top reaction you get the following (see below). AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10 Ag+ + 2 NH3(aq) ↔ Ag(NH3)2
+ Keq = 6.25x10-8 AgCl + 2 NH3(aq) ↔ Ag(NH3)2
+ + Cl- K = 0.1125 This means that all the silver in solution will end up being Ag(NH3)2
+. If you dissolve all of the AgCl in 1 liter of solution, the concentration of Ag(NH3)2
+ must be,
100 g AgCl/143.35 g/mol = 0.6976 mole AgCl => 0.6976 mole Ag+ becomes, 0.6976 mole Ag(NH3)2
+ and 0.6976 mole Cl- in 1 liter of solution. So, [Ag(NH3)2
+] = 0.6976 M and [Cl-] = 0.6976 M Solving for the concentration of NH3(aq),
1125.0][NH
M] [0.6976 M] .69760[
][NH
][Cl ])[Ag(NH2
32
3
23 ==
−+
[NH3] = 2.08 M
30) Copper(I) ions in aqueous solution react with NH3 according to, Cu+ + 2 NH3 ↔ Cu(NH3)2
+ Keq = 6.3x1010 Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium concentration of NH3 is 0.185 M.
CuBr ↔ Cu+ + Br- Ksp = 5.3x10-9 Cu+ + 2 NH3 ↔ Cu(NH3)2
+ Keq = 6.3x1010 CuBr + 2 NH3 ↔ Cu(NH3)2
+ + Br- K = 333.9 For every Cu(NH3)2
+ there must be one Br-, so [Cu(NH3)2+] = [Br-] = x
9.333[0.185]][NH 2
2
23
2
== xx x = 3.38 M CuBr
31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-). PbS <---> Pb2+ + S2- Ksp = 8x10-28
H2S <---> 2 H+ + S2- Keq = 1x10-20
Note: the Pb2+ concentration can only be as high as the least soluble compound so the solubility of PbS will determine the concentration of Pb2+
[H+]2 [S2-]/[H 2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1]2 = 1x10-21 M S2-
Substitute this result into the following equation,
[Pb2+] [S2-] = 8x10-28 [Pb2+] = 8x10-28/[S2-] = 8x10-28/[1x10-21] = 8x10-7 M
Now, all equilibriums must be satisfied simultaneously. Since this is the maximum concentration of Pb2+ that can be in solution, we can use this value to calculate the Cl-
concentration from the PbCl2. PbCl2(s) ↔ Pb2+ + 2 Cl- [Pb2+] [Cl -]2 = 1.6x10-5 => [8x10-7 M] [Cl -]2 = 1.6x10-5 [Cl-] = 4.47 M This concentration of Cl- is fairly high. It can only be this high if all of the PbCl2 dissolved. So, if you put H2S in a solution of PbCl2, all of the PbCl2 would dissolve and become PbS.
32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6. Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-).
[H+]2 [S2-]/[H 2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-6]2 = 1x10-9 M S2-
Substitute this result into the following equation,
Fe2S3(s) ↔ 2 Fe3+ + 3 S2-
[Fe3+]2 [S2-]3 = 1.6x10-21 [Fe3+] = (1.6x10-21/[S2-]3)½ = (1.6x10-21/[1x10-9]3)½ = 1264.9 M Fe3+ => 632.45 M Fe2S3
33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH = 3? CdS <--> Cd2+ + S2- Ksp = 8x10-27 H2S <--> 2H+ + S2- Keq = 1x10-20
[H+]2 [S2-]/[H 2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2-
Substitute this result into the following equation,
[Cd2+] [S2-] = 8x10-27 [Cd2+] = 8x10-27/[S2-] = 8x10-27/[1x10-17] = 8x10-12 M
34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated solution of H2S = 0.10 M at pH = 3? Ag2S <--> 2 Ag+ + S2- Ksp = 1.6x10-49 H2S <--> 2H+ + S2- Keq = 1.1x10-20
[H+]2 [S2-]/[H 2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2-
Substitute this result into the following equation,
[Ag+]2 [S2-] = 1.6x10-49 [Ag+] = 1.6x10-49/[S2-] = 1.6x10-49/[1x10-17] = 1.6x10-32 M Ag+
35) How much of each precipitate will form and what will the concentration of lead be if 0.075 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M Mg(NO3)2 and 0.5 M Pb(NO3)2? MgOx(s) ↔ Mg2+ + Ox2- Ksp = 8.6x10-5 PbOx(s) ↔ Pb2+ + Ox2- Ksp = 4.8x10-10
Neither compound is very soluble so you can begin by assuming that 100% of the Pb2+ and Mg2+ will react to form solids.
(0.30 M)(0.100 L) = 0.03 mole Mg(NO3)2 (0.50 M)(0.100 L) = 0.05 mole Pb(NO3)2 Start with the least soluble compound (PbOx), You have 0.075 mole Ox2- and 0.50 mole of Pb2+, so all of the Pb2+ will react to make 0,50 mole of PbOx and leave 0.025 mole of Ox2- in solution. The remaining 0.025 mole of Ox2- will react with the 0.03 mole Mg(NO3)2 leaving 0.005 mole of Mg2+ in solution and no Ox2-. The [Mg2+] = (0.005 mol)(0.100 L) = 0.05 M Mg2+. [Mg2+] [Ox2-] = [0.05 M] [Ox2-] = 8.6x10-5 [Ox2-] = 1.72x10-3 M [Pb2+] [Ox2-] = [Pb2+] [1.72x10-3 M] = 4.8x10-10 [Pb2+] = 2.79x10-7 M
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 9? Ag2CO3 <---> 2 Ag+ + CO3
2- Ksp = 8.1x10-12 H2CO3 <--> 2 H+ + CO3
2- Ka = 2.02x10-17
M 0.6868][CO2.02x100.034M
)(COM)(1x10
]CO[H
][CO][H 23
1723
29
32
23
2
=⇒== −−−−−+
[Ag+]2 [CO3
2-] = [Ag+]2 [0.6868 M] = 8.1x10-12 [Ag+] = 3.43x10-6 M Solubility of Ag2CO3 = (3.43x10-6 M)/2 = 1.72x10-6 M Ag2CO3
Chemistry 121 Name_____________ Second Exam April 18, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator.
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1) Which of the following compounds is the most soluble in water? Circle one. Show your work. BaSO4 Ksp = 1.6x10-10 PbI2 Ksp = 1.2x10-13 Ag3PO4 Ksp = 1x10-22 2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of concentration. I wanted to know the concentration so I measured the pH and found it to be pH = 2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts of 0.8 M NaOH H3AsO4 <---> H2AsO4
- + H+ Ka1= 5.62x10-5 H2AsO4
- <---> HAsO42- + H+ Ka2= 1.7x10-8
HAsO42- <---> AsO4
3- + H+ Ka3= 3.95x10-12 a) 0 mL b) 150 mL c) 225 mL d) 400 mL e) 450 mL f) 600 mL f) If you wanted to titrate to the first endpoint which indicator would you use? Circle one. Indicator Range Methyl Red 4.8 - 6.0 Bromothymol Blue 6.2 - 7.6 Cresol Red 7.0 - 8.8 Phenolphthalein 8.2 - 10
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of pH = 11.6? 4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 8? MgCO3 <---> Mg2+ + CO3
2- Ksp = 8.1x10-12 H2CO3 <--> 2 H+ + CO3
2- Ka = 2.02x10-17 5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example.
Chemistry 121 Name_____________ Second Exam April 18, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator.
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1) Which of the following compounds is the most soluble in water? Circle one. Show your work. BaSO4 Ksp = 1.6x10-10 (x)(x) = x2 = 1.6x10-10 x = 1.26x10-5 PbI2 Ksp = 1.2x10-13 (x)(2x)2 = 4x3 = 1.2x10-13 x = 3.11x10-5 Ag3PO4 Ksp = 1x10-22 (3x)3(x) = 27x4 = 1x10-22 x = 1.39x10-6 2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of concentration. I wanted to know the concentration so I measured the pH and found it to be pH = 2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
-log[H+] = 2.87 [H+] = 1.35x10-3 M also, when HAc dissociates, [H+] = [Ac-] HAc ↔ H+ + Ac- x 1.35x10-3 1.35x10-3 x = 0.1013 M HAc
( ) 5
23-
106.1x
1.35x10 −= x
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts of 0.8 M NaOH H3AsO4 <---> H2AsO4
- + H+ Ka1= 5.62x10-5 pKA1 = 4.25 H2AsO4
- <---> HAsO42- + H+ Ka2= 1.7x10-8 pKA2 = 7.74
HAsO42- <---> AsO4
3- + H+ Ka3= 3.95x10-12 pKA3 = 11.40
a) 0 mL 52
1062.52.1
−=−
xx
x x = 8.21x10-3 pH = 2.09
b) 150 mL 2
7.744.252
pKpKpH A2A1 +=+= pH = 5.995
c) 225 mL pH = 7.74 d) 400 mL 100 mL into 3rd Region (0.8 M)(0.100 L) = 0.08 mol OH- pH = 11.70
e) 450 mL 312-
-1434 1053.2
3.95x101x10
Kb and 0.2182M0.550L0.12mol
][AsO −− ==== x
x = 7.43x10-2 M OH- pH = 12.87 f) 600 mL (0.150L)((0.80M) = 0.12 mol OH-
−−
= OH M 0.1714L 0.700OH mol 0.12
pH = 13.23
g) If you wanted to titrate to the first endpoint which indicator would you use? Circle one. Indicator Range Methyl Red 4.8 - 6.0 Bromothymol Blue 6.2 - 7.6 since pH = 5.995 @ first endpoint Cresol Red 7.0 - 8.8 choose an indicator that changes Phenolphthalein 8.2 – 10 just after 5.995
M1V1 = M2V2 (1.2M)((0.100L) = (0.08 M)(V2) V2 = 150 mL for each region Also (1.2M)(0.100L) = 0.12 mol H3AsO4
0.080.120.08
log11.40pH−
+=
32
1054.22182.0
−=−
xx
x
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of pH = 11.6?
The pH is in the 3rd region so you will have to add 300 mL of NaOH to your answer,
x
x
−+=
12.0log4.116.11 x = 0.7357 mol OH-
0.7357 mol OH- = (0.8 M)(V) V = 91.97 mL + 300 mL = 391.97 mL total
4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 8? MgCO3 <---> Mg2+ + CO3
2- Ksp = 8.1x10-12 H2CO3 <--> 2 H+ + CO3
2- Ka = 2.02x10-17
M6.868x10][CO2.02x100.034M
)M)(CO(1x10]CO[H
]][CO[H 323
1723
8
32
23 −−−
−−−+
=⇒==
[Mg2+] [CO3
2-] = [Mg2+] [6.868x10-3 M] = 8.1x10-12 [Mg2+] = 1.18x10-9 M
5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example. Amphoterism = the ability of a metal ion to be soluble in acidic and basic solutions
Con
cent
ratio
n of
Met
al Io
n
pH0 7 14
Chemistry 121 Name ________________ Second Exam April 10, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator.
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1) Which of the following compounds is the most soluble in water? Show your work. BaF2 Ksp = 1.84x10-9 Mg3(PO4)2 Ksp = 1.04×10-15
Ag3PO4 Ksp = 1.2x10-15 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HAc Na3PO4 + HCl HCl + KOH
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of 0.6 M NaOH
H2CO3 <---> H+ + HCO3- Ka1= 4.3x10-7
HCO3- <---> H+ + CO3
2- Ka2= 4.7x10-11 a) 0 mL b) 100 mL c) 150 mL d) 200 mL e) 300 mL f) 365 mL g) 400 mL h) 450 mL i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH = 10?
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3? Ag2S <---> 2 Ag+ + S2- Ksp = 6.0x10-50 H2S <---> 2 H+ + S2- Keq = 6.84x10-23 5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and FeS.
5b) Define amphoterism and give an example.
Chemistry 121 Name Second Exam April 10, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator.
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1) Which of the following compounds is the most soluble in water? Show your work. BaF2 Ksp = 1.84x10-9 (x)(2x)2 = 4x3 = 1.84x10-9 x = 7.72x10-4 Mg3(PO4)2 Ksp = 1.04×10-15 (3x)3(2x)2 = 108x5 = 1.04x10-15 x = 3.95x10-4 most
soluble
Ag3PO4 Ksp = 1.2x10-15 (3x)3 (x) = 27x4 = 1.2x10-15 x = 8.16x10-5 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HAc H2O + Ac- –> HAc + OH- Na3PO4 + HCl HPO4
2- –> H+ + PO43-
HCl + KOH H2O –> H+ + OH-
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of 0.6 M NaOH
H2CO3 <---> H+ + HCO3- Ka1= 4.3x10-7
HCO3- <---> H+ + CO3
2- Ka2= 4.7x10-11 a) 0 mL x2/(1.5-x) = 4.3x10-7 x = 8.03x10-4M H+ pH = 3.095 b) 100 mL ½ way point so pH = 6.37 c) 150 mL - use H-H equation (0.6M)(0.150L) = .09mole NaOH and (1.5M)(0.080L) = 0.12mole H2CO3 pH = 6.37 + log(0.09/(0.12-0.09)) pH= 6.847 d) 200 mL first end point so pH = (pka1 + pka2)/2 –> pH = (6.37 + 10.33)/2 = 8.35 e) 300 mL 2nd midpoint so pH = pKa2 = 10.33 f) 365 mL - you are now 165 mL into the second region - use H-H equation (0.06M)(0.165L) = 0.099 mole NaOH and you still have 0.12 mole of acid (HCO3
-) pH = 10.33 + log(0.099/(0.12-0.099)) pH = 11.00 g) 400 mL - at final endpoint pKb = Kw/Ka2 = 1x10-14/4.7x10-11 = 2.13x10-4 also [CO3
2-] = 0.12 mole/ (0.480L total) = 0.25M CO32-
x2/(0.25 - x) = 2.13x10-4 –> x = 0.007297 M OH- pH = 11.86 h) 450 mL - this is 50 mL past the final endpoint
(0.050L)(0.60M) = 0.030 mole excess OH- in a total of 530 mL of solution. Therefore, [OH-] = 0.030 mole / 0.530 L = 0.0566 M OH- –> pH = 12.75
i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH = 10? The pH is in the 2nd region so you must add 200 mL to start this titration. Use H-H,
10 = 10.33 + log(x/(0.12-x)) –> x = 0.0382 mole NaOH 0.0382 mole NaOH = MV = (0.80M)(V) –> V = 0.0478 L or 47.8 mL, therefore, 200 mL NaOH to get to first endpoint + 47.8 mL = 247.8 mL of 0.80 M NaOH
0 7 14
[Ions
]
pH
CuS FeS
At this pH FeS is highly solubleand CuS is not very soluble.
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3?
Ag2S <---> 2 Ag+ + S2- Ksp = 6.0x10-50 H2S <---> 2 H+ + S2- Keq = 6.84x10-23
[H+]2 [S2-]/[H2S] = 6.84x10-23
[S2-] = 6.84x10-23 [H2S]/[H+]2 = 6.84x10-23 [0.01]/[1x10-3]2 = 6.84x10-19 M S2-
Substitute this result into the following equation,
[Ag+]2 [S2-] = 6x10-50 [Ag+]2 = 6x10-50/[S2-] = 6x10-50/[6.84x10-19] = 8.77x10-32
taking the square root of this answer to find [Ag+], [Ag+] = 2.96x10-16 M Ag+ 5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and FeS. 5b) Define amphoterism and give an example.
Amphoterism is the ability of an ion to be soluble in either an acid or a base but not in between. This property is commonly exhibited with transition metals like Zinc,
Zn(OH)2(s) –> Zn2+ + 2 OH- soluble in acid Zn(OH)2(s) + 2 OH- –> Zn(OH)4
2- soluble in base
