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Equilibrium Problems
CHE
Sherman Henzel
ICE Tables To ICE or not to ICE, that is the question. ICE tables are used when initial (I) concentrations are given. If only equilibrium concentrations are given, then ICE tables are not employed. ICE problems include pH determination of weak acids and bases, equilibrium concentrations, K determinations, reacting species such as buffer reactions and salts that can form a precipitate, titrations of weak acids and based, solubility problems, and precipitation problems. Examples:
- Determine the pH of a solution that is 0.1500 M in HC 2 H 3 O 2. 0.1500 M HC 2 H 3 O 2 is an initial concentration and an ICE table is employed. answer: pH = 2. Similar problems: Chapter 18 pages 11, 12, 15 , and 17.
- An organic acid HOrg (MW = 325.00 g/mol) is soluble in water to the extent of 1.58 g HOrg/L. If the pH of a 500.0 mL HOrg solution has a pH of 4.821, what is the Ka for this acid? HOrg + H 2 O ( Org -^ + H 3 O+ The pH is measured at equilibrium, but the solubility of the acid is an initial concentration and an ICE table is employed. answer: Ka = 4.70 x 10- Similar problems: Chapter 18 pages 13 and 14.
- Calculate the pH of a solution that is 0.1250 M in hydroxylamine. 0.1250 M hydroxylamine is an initial concentration and an ICE table is employed. answer: pH = 9. Similar problems: Chapter 18 page 18.
- Calculate the pH of a solution that is 0.1150 M in NaOCl. 0.1150 M NaOCl is an initial concentration and an ICE table is employed. answer: pH = 10. Similar problems: Chapter 18 pages 20, 21, 29, 30, and 31.
Equilibrium Problems page 2
- Calculate the [H 2 S]e, [HS-^ ]e, [S-2^ ]e and [H 3 O+^ ]e in a solution that is 0.1750 M in H 2 S. 0.1750 M H 2 S is an initial concentration and an ICE table is employed. answer: [H 2 S] = 0.1750 M, [HS-^ ] = 1.4 x10 -4^ M, [H 3 O+^ ] = 1.4 x 10 -4^ M, [S-2^ ] = 1.0 x 10 - Similar problems: Chapter 18 pages 23 and 24.
- Determine the pH in a solution that is 0.1500 M in HNO 2 and 0.1250 M in HCl. 0.1250 M 0.1500 M HNO 2 and 0.1250 M HCl are both initial concentrations and an ICE table is employed. answer: pH = 0. Similar problems: Chapter 19 page 2.
- Calculate the pH of a solution that is 0.5000 M in HC 2 H 3 O 2 and 0.2500 M in NaC 2 H 3 O 2. 0.5000 M HC 2 H 3 O 2 and 0.2500 M NaC 2 H 3 O 2 are both initial concentrations and an ICE table is employed. answer: pH = 4. Similar problems: Chapter 19 page 3.
- Calculate the pH of a buffer that is 0.1025 M in HONH 2 and 0.2000 M HONH 3 Cl. 0.1025 M HONH 2 and 0.2000 M HONH 3 Cl are both initial concentrations and an ICE table is employed. answer: pH = 5. Similar problems: Chapter 19 pages 5 and 8 (bottom).
- What is the effect of adding 0.01000 moles of NaOH to 1.00 L of the buffer in problem 8 above? 0.01000 M NaOH, a strong base, will react with the conjugate acid of the buffer and an ICE table is employed. answer: DpH = 0.06 pH units Similar problems: Chapter 19 pages6, 7, 8, 9 and 10.
Equilibrium Problems page 4
- What is the solubility of FeS in 1.050 M HCl? Solubility problems employ ICE tables. answer: S = 0.5 21 M Similar problems: Chapter 19 pages 66 and 67.
No ICE Tables ICE tables are not employed when equilibrium (E) concentrations are given.. Non-ICE problems include pH determination of strong, buffer preparation, titrations of strong acids and bases, post equivalence point for weak acid and base titrations, Ksp determination, calculation of Q, ion separation by precipitation and preventing precipitation.
- Calculate the [H 3 O+^ ], [OH-^ ] and [Cl -^ ] in 500.0 mL of a solution that is 0.1115 M in HCl. Strong acid does not employ and ICE table. answer: [H 3 O+^ ] = [NO- 3 ] = 0.1115 M, [OH-^ ] = 9.0 x 10 -14^ M Similar problems: Chapter 18 pages 7 and 8.
- What mass of NaNO 2 must be added to 250.0 mL of 0.350 M HNO 2 in order to produce a buffer with a pH of 3.50? Buffer preparation involves all concentrations at equilibrium and does not employ ICE tables. answer: 9.7 f NaNO 2 Similar problems: Chapter 19 page 4.
- Describe the preparation of 250.0 mL of a buffer with a pH of 5.000 from 0.1500 M HC 2 H 3 O 2 and 0.2000 M NaC 2 H 3 O 2. Buffer preparation involves all concentrations at equilibrium and does not employ ICE tables. answer: Mix 108 mL of 0.1500 M HC 2 H 3 O 2 with 142 mL of 0.2000 M NaC 2 H 3 O 2. Similar problems: Chapter 19 pages 12 and 13.
Equilibrium Problems page 5
- 25.00 mL of 0.1500 M HCl are titrated with 0.3000 M NaOH. Calculate the pH after 7. mL of NaOH are added. Strong acid and strong base titrations do not employ an ICE table. answer: pH = 1. Similar problems: Chapter 19 pages 15 to 20.
- 50.00 mL of 0.2000 M HC 2 H 3 O 2 are titrated with 0.4000 M NaOH. Calculate the pH after 30.00 mL of NaOH are added. Weak acid strong base tirations post equivalence point do not employ ICE tables. answer: pH = 12. Similar problems: Chapter 19 pages 25 and 29.
- A 125.0 mL sample is removed from a water solution saturated with Ca(OH) 2 at 25.0 C. The water is completely evaporated from the solution and a deposit of 0.1030 g of Ca(OH) 2 is obtained. Calculate the Ksp of Ca(OH) 2 at 25.0 C. In a saturated solution all concentrations are at equilibrium and an ICE table is not employed. answer: Ksp = 5.502 x 10- Similar problems: Chapter 19 page 36.
- It is desired to separate Ba 2+^ from Ag+^ by the slow addition of SO2- 4 to a solution in which the concentration of Ba 2+^ is 0.002500 M and the concentration of Ag +^ is 0.0001250 M. (a) Which ion will precipitate first and at what [SO2- 4 ]? (b) What will be the concentration of the first ion to precipitate when the second ion just begins to precipitate? (c) Is the separation effective? Separation problems involve equilibrium concentrations and ICE stables are not employed. answer: (a) Ba 2+^ will precipitate first when [SO2- 4 ] = 4.4 x 10 -8^ M (b) [Ba2+^ ] = 1.2 x 10 -13^ M (c) Yes. Similar problems: Chapter 19 pages 40 and 41.
Equilibrium Problems page 7
Equilibrium Problems Key
CHE
Sherman Henzel
ICE Tables Examples:
- Determine the pH of a solution that is 0.1500 M in HC 2 H 3 O 2.
HC 2 H 3 O 2 + H 2 O (^ C 2 H 3 O- 3 + H 3 O+ I 0.1500 0 0 C -x +x +x Ex 0.1500 - x x x E =0. 2 3 2 3 5 2 2 3 2
[C H O ][H O ] 1.74 x 10 x ; x 0.
[HC H O ] 0.
− + −
[H 3 O+^ ] = 0.00162 M; pH = 2.
- An organic acid HOrg (MW = 325.00 g/mol) is soluble in water to the extent of 1.58 g HOrg/L. If the pH of a 500.0 mL HOrg solution has a pH of 4.821, what is the Ka for this acid? HOrg + H 2 O ( Org -^ + H 3 O+
[Horg]I = 1.58 g HOrgL x325.00 g HOrg1 mol HOrg^ =0.00486 mol HOrgL
pH = 4.821; [H 3 O+^ ]E = 1.51 x 10-5^ M
HOrg + H 2 O ( Org -^ + H 3 O+ I 0.00486 0 0 C -x +x +x Ex 0.00486-x x x E 0.00485 1.51 x 10-5^ 1.51 x 10- 3 5 2 8
K a [Org ][H O ][HOrg] (1.51 x 100.00485^ ) 4.70 x 10
− + (^) − −
Equilibrium Problems Key page 8
- Calculate the pH of a solution that is 0.1250 M in hydroxylamine.
HONH 2 + H 2 O ( HONH+ 3 + OH- I 0.1250 0 0 C -x +x +x Ex 0.1250 - x x x E =0. 3 9 2 5
K b [HONH ][OH ][HONH ] 2 9.1 x 10 0.1250x ; x 3.4 x 10
[OH-^ ] = 3.4 x 10 -5^ M; pOH = 4.47; pH = 9.
- Calculate the pH of a solution that is 0.1150 M in NaOCl.
OCl -^ + H 2 O (^ HOCl + OH- I 0.1150 0 0 C -x +x +x Ex 0.1150 - x x x E =0. w^147 b (^) a 8 (^27 )
K [HOCl][OH ]^ K 1.00 x 10 3.39 x 10
[OCl ] K 2.95 x 10
x 3.39 x 10 ; x 1.97 x 10
− − − − − − −
[OH-^ ] = 1.97 x 10 -4^ M; pOH = 3.70; pH = 10.
Equilibrium Problems Key page 10
- Calculate the pH of a solution that is 0.5000 M in HC 2 H 3 O 2 and 0.2500 M in NaC 2 H 3 O 2.
HC 2 H 3 O 2 + H 2 O ( C 2 H 3 O- 3 + H 3 O+ I 0.5000 0.2500 0 C -x +x +x Ex 0.5000 - x 0.2500 + x x E =0.5000 =0. 2 3 2 3 5 5 2 3 2
[C H O ][H O ] 1.74 x 10 0.2500x; x 3.48 x 10
[HC H O ] 0.
− + − −
[H 3 O+^ ] = 3.48 x 10 -5^ M; pH = 4.
- Calculate the pH of a buffer that is 0.1025 M in HONH 2 and 0.2000 M HONH 3 Cl.
HONH 2 + H 2 O (^ HONH+ 3 + OH- I 0.1025 0.2000 0 C -x +x +x Ex 0.1025 - x 0.2000 + x x E =0.1025 =0. b^39 2
K [HONH ][OH ]^ 9.1 x 10 0.2000x; x 4.7 x 10
[HONH ] 0.
[OH-^ ] = 4.7 x 10 -9^ M; pOH = 8.33; pH = 5.
- What is the effect of adding 0.01000 moles of NaOH to 1.00 L of the buffer in problem 8 above? HONH+ 3 + OH-^ ( HONH 2 + H 2 O I 0.2000 0.01000 0. C -0.01000 -0.01000 +0. I 0.1900 0 0. C +x +x -x Ex 0.1900 + x x 0.1125 - x E =0.1900 =0. b^39 2
K [HONH ][OH ]^ 9.1 x 10 0.1900x; x 5.4 x 10
[HONH ] 0.
[OH-^ ] = 5.4 x 10 -9^ M; pOH = 8.27; pH = 5.73; DpH = 5.73 - 5.67 = 0.06 pH units
Equilibrium Problems Key page 11
- 50.00 mL of 0.2000 M HC 2 H 3 O 2 are titrated with 0.4000 M NaOH. Calculate the pH after 15.00 mL of NaOH are added.
[HC H O ] 2 3 2 0.2000 M 1 x 50.00 mL65.00 mL 0.1538 M
[OH ] 0.4000 M^ x 15.00 mL 0.09231 M
1 65.00 mL
−
HC 2 H 3 O 2 + OH-^ ( C 2 H 3 O- 3 + H 2 O
I 0.1538 0.09231 0
C -0.09231 -0.09231 +0.
I 0.0615 0 0.
C +x +x -x Ex 0.0615 + x 0.09231 - x E =0.0615 =0. 2 3 2 a^1410
K b [HC H O ][OH ][C H O ] 2 3 2 KK^ a 1.00 x 101.74 x 10^5 0.0615x0.09231; x 8.63 x 10
− − −
[OH-^ ] = 8.63 x 10 -10^ M; pOH = 9.06; pH = 4.
- Calculate the solubility of AgCl in water.
AgCl (s) ( Ag+^ + Cl - I 0 0 C +x +x Ex x x E Ksp = [Ag+^ ][Cl -^ ] = 1.8 x 10 -10^ = x 2 ; x = 1.3 x 10- S = 1.3 x 10-5^ M
Equilibrium Problems Key page 13
- Calculate the solubility of Ag 2 CO 3 in a solution buffered to a pH of 3.00. Ag 2 CO 3 (s) ( 2 Ag +^ + CO-2 3 Ksp = 8.1 x 10- CO-2 3 + H 3 O+^ ( HCO- 3 + H 2 O 1/Ka2 = 1/5.6 x 10 -11^ = 1.8 x 10^10 HCO- 3 + H 3 O+^ ( H 2 CO 3 + H 2 O 1/Ka1 = 1/4.2 x 10 -7^ = 2.4 x 10^6 Ag 2 CO 3 (s) + 2 H 3 O+^ ( 2 Ag+^ + H 2 CO 3 + 2 H 2 O K = 8.1 x 10 -12^ x1.8 x 10^10 x 2.4 x 10^6 = 3.5 x 10^5 pH = 3.00; [H 2 O+^ ] = 1.0 x 10 -3^ M Ag 2 CO 3 (s) + 2 H 3 O+^ ( 2 Ag +^ + H 2 CO + 2 H 2 O I 1.0 x 10 -3^0 C ------ +2x +x Ex ------ 2x x E 1.0 x 10- (^2 2 352 ) 3 2 3 2 6
K [Ag ] [H CO ]^ 3.5 x 10 (2x) x^ 4x ; x 0.
[H O ] (1.0 x 10 ) 1.0 x 10
S = 0.44 M
- What is the molar solubility of AgCl in 0.1015 M NH 3? AgCl (s) ( Ag +^ + Cl -^ Ksp = 1.8 x 10- Ag +^ + 2 NH 3 ( Ag(NH 3 ) + 2 Kf = 1.6 x 10^7 AgCl (s) + 2 NH 3 ( Ag(NH 3 ) + 2 + Cl -^ K = 1.8 x 10 -10^ x 1.6 x 10^7 = 0. AgCl (s) + 2 NH 3 ( Ag(NH 3 ) + 2 + Cl - I 0.1015 0 0 C -2x +x +x Ex 0.1015 -2x x x 3 2 2 3 2 2 (^23) 2
K [Ag(NH ) ][Cl ]^ 0.0029 x
[NH ] (0.1015 2x)
x 0.0029; x 0.053; x 4.9 x 10
(0.1015 2x) 0.1015 2x
−
− =^ − =^ =
S = 4.9 x 10-3^ M
Equilibrium Problems Key page 14
- What is the solubility of FeS in 1.050 M HCl? FeS (s) ( Fe 2+^ + S2-^ Ksp = 6 x 10 - S-2^ + H 3 O+^ ( HS-^ + H 2 O K = 1/K (^) a2 = 1/1.0 x 10 -14^ = 1.0 x 10^14 HS-^ + H 3 O+^ ( H 2 S + H 2 O K = 1/K (^) a2 = 1/1.1 x 10 -7^ = 9.1 x 10^6 FeS (s) + 2 H 3 O+^ ( Fe 2+^ + H 2 S K = 6 x 10 -18^ x 1.0 x 10^14 x 9.1 x 10^6 = 5. 46 x 10^3 FeS (s) + 2 H 3 O+^ ( Fe 2+^ + H 2 S + 2 H 2 O I 1.050 0 0 C -2x +x +x Ex 1.050 - 2x x x (^2 ) 3 2 46 2 3 2 1 46 2 39 21
K [Fe^ ][H S]^ 5. x 10 x
[H O ] (1.050 2x)
5. x 10 x^ ; 7. x 10 x ; x 0.
(1.050 2x) 1.050 2x
S = 0.5 21 M
Non-ICE Tables Examples:
- Calculate the [H 3 O+^ ], [OH-^ ] and [NO- 3 ] in 500.0 mL of a solution that is 0.1115 M in HNO 3. HNO 3 + H 2 O ) H 3 O+^ + NO- 3 [H 3 O+^ ][OH-^ ] = 1.00 x 10 -14; 0.1115[OH-^ ] = 1.00 x 10 -14; [OH-^ ] = 9.0 x 10 -14^ M [H 3 O+^ ] = [NO- 3 ] = 0.1115 M, [OH-^ ] = 9.0 x 10 -14^ M
- What mass of NaNO 2 must be added to 250.0 mL of 0.350 M HNO 2 in order to produce a buffer with a pH of 3.50? pH = 3.50; [H 3 O+^ ] = 3.2 x 10 -4^ M 4
a 2 3 4 2 2 2
K [NO ][H O ]^ 5.13 x 10 [NO ] 3.2 x 10 ; [NO ] 0.56 M
[HNO ] 0.
− + (^) − − − −
(^2 ) 2
0.56 mol NaNO x 68.996 g NaNO x 0.2500 L 9.7 g NaNO
L 1 mol NaNO 1 =
Equilibrium Problems Key page 16
- A 125.0 mL sample is removed from a water solution saturated with Ca(OH) 2 at 25.0 C. The water is completely evaporated from the solution and a deposit of 0.1030 g of Ca(OH) 2 is obtained. Calculate the Ksp of Ca(OH) 2 at 25.0 C.
(^2 ) 2 2 2 2 2 2 2
0.1030 g Ca(OH) x 1 mol Ca(OH) 0.001390 mol Ca(OH)
1 74.093 g Ca(OH)
0.001390 mol Ca(OH) x 1 mol Ca x 1 0.01112 M Ca
1 1 mol Ca(OH) 0.1250 L
0.001390 mol Ca(OH) x 2 mol OH x 1 0.02224 M OH
1 1 mol Ca(OH) 0.1250 L
− −
Ksp = [Ca2+^ ][OH-^ ]^2 = 0.01112 x (0.02224) 2 = 5.502 x 10-
- It is desired to separate Ba 2+^ from Ag+^ by the slow addition of SO2- 4 to a solution in which the concentration of Ba 2+^ is 0.002500 M and the concentration of Ag +^ is 0.0001250 M. (a) Which ion will precipitate first and at what [SO2- 4 ]? (b) What will be the concentration of the first ion to precipitate when the second ion just begins to precipitate? (c) Is the separation effective? BaSO 4 (s) ( Ba 2+^ + SO2- 4 Ksp = 1.1 x 10- Ag 2 SO 4 (s) ( 2 Ag+^ + SO2- 4 Ksp = 1.4 x 10- (a) Ksp = [Ba2+^ ][SO2- 4 ] = 1.1 x 10 -10^ = 0.002500 [SO2- 4 ]; [SO2- 4 ] = 4.4 x 10 -8^ M Ksp = [Ag+^ ]^2 [SO2- 4 ] = 1.4 x 10 -5^ = (0.0001250) 2 [SO2- 4 ]; [SO2- 4 ] = 896 M Ba 2+^ will precipitate first when [SO2- 4 ] = 4.4 x 10 -8^ M (b) [Ba2+^ ] 896 = 1.1 x 10 -10; [Ba 2+^ ] = 1.2 x 10 -13^ M (c) Yes.
Equilibrium Problems Key page 17
- Eight drops of 0.01500 M NaCl are added to 175.0 mL of 0.007500 M AgNO 3. Will AgCl precipitate? 20 drops equal 1 mL.
5
3 3
8 d x 1 mL 0.40 mL
1 20 d
0.01500 mol NaCl x 0.40 mL x 1 mol Cl 3.4 x 10 M Cl
L 175.4 mL 1 mol
0.007500 mol AgNO x 175.0 mL x 1 mol Ag 0.007483 M Ag
L 175.4 mL 1 mol AgNO
− − −
Q = [Ag+^ ][Cl -^ ] = 3.4 x 10 -5^ x 0.007483 = 2.6 x 10- Q > Ksp therefore AgCl (s) will form
- What concentration of C 2 H 5 NH+ 3 must be maintained in order to prevent the precipitation of Mg(OH) 2 in a solution that is 0.01250 M in Mg(NO 3 ) 2 and 0.003500 M in C 2 H 5 NH 2? Ksp = [Mg+2][OH-^ ]^2 = 1.8 x 10-11^ = 0.01250[OH-^ ]^2 ; [OH-^ ] = 3.8 x 10 -5^ M
2 5 3 4 2 5 3 5
K b [C H NH ][OH ][C H NH ] 2 5 2 4.4 x 10 [C H NH ]3.8 x 100.003500 ; [C H NH ] 2 5 3 0.041 M
- What is the minimum concentration of NH 3 required to prevent AgCl from precipitating in a solution that is 0.0002500 M in AgNO 3 and 0.00009500 M in NaCl? Ksp = [Ag+^ ][Cl -^ ] = 1.8 x 10 -10^ = [Ag+^ ]0.00009500; [Ag+^ ] = 1.9 x 10 -6^ M
f (^3 22 76 ) 3 3 3 3
K [Ag(NH ) ]^ 1.6 x 10 0.
[Ag ][NH ] 1.9 x 10 [NH ]
[NH ] 2.9 x 10 M
0.0002500 mol Ag x 2 mol NH 3 0.0005000 M NH 3
L 1 mol Ag
[NH 3 ] = 0.0029 M + 0.0005000 M = 0.0034 M
