AP Chemistry Unit 7- Homework Problems Equilibrium and K sp

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1 AP Chemistry Unit 7- Homework Problems Equilibrium and K sp Nature of the Equilibrium State 1. Draw on this graph where equilibrium has been reached. [X] equilibrium time 2. What are three qualities of any equilibrium equation? a. Reversible b. Dynamic c. Occur when there are a stable ratio of products: reactants 3. For a general equation: aa + bb cc + dd, write the equation for K c. K c = [C] c [D] d [A] a [B] b Developing K eq 1. For each of the equations below, write the expression for K c : a. 2 H 2 S (g) 2 H 2 (g) + S 2 (g) K c = [H 2 ] 2 [S 2 ] [H 2 S] 2 b. HCN (aq) + H 2 O (l) H 3 O + (aq) + CN -1 (aq) K c = [H 3 O + ] [CN -1 ] [HCN] c. PbCl 2 (s) Pb +2 (aq) + 2 Cl -1 (aq) K c = [Pb +2 ] [Cl -1 ] 2

2 2. For each of the equations below, write the expression for K p : a. SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K p = P SO2 P Cl2 P SO2Cl2 b. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) K p = P CO2 P H2 P CO P H2O c. C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O (g) K p = P 6 CO2 P 6 H2O P 6 O2 3. Put the following K values in order of increasing product-favored ability. a. K = 4x10-5 b. K = 2x10-9 c. K = 7x10-5 d. K = 3x10-3 b < a < c < d Equilibrium Mathematics 1. The equation: C (s) + 2 H 2 O (g) CO (g) + H 2 (g) has a value of K c = 2.5 x10-6 What is the value of K c for: CO (g) + H 2 (g) C (s) + 2 H 2 O (g)? K c = (2.5x10-6 ) -1 = 4x10 5 What is the value of K c for: 2 C (s) + 4 H 2 O (g) 2 CO (g) + 2 H 2 (g) K c = (2.5x10-6 ) 2 = 6.25x The equation: H 2 O (g) H 2 (g) + ½ O 2 (g) has a value of K p = 4.9 x10-3 What is the value of K p for: 2 H 2 O (g) 2 H 2 (g) + O 2 (g)? K p = (4.9x10-3 ) 2 = 2.4x10-5 What is the value of K p for: H 2 (g) + ½ O 2 (g) H 2 O (g)? K p = (4.9x10-3 ) -1 = The equation: 2 NH 3 (g) N 2 (g) + 3 H 2 (g) has a value of K c = 2.7 x10-4 At STP, What is the value for K p? K p = K c (RT) n = (2.7 x10-4 )[(0.0821)(273 K)] 2 = At STP, What is the value of K c for: K c = (2.7 x10-4 ) -1/2 = 60.9 ½ N 2 (g) + 3/2 H 2 (g) NH 3 (g)? 4. The equation: N 2 O 4 (g) 2 NO 2 (g) has a value of K p = 3.2x10-4 What is the value for K c at 300 K?? K p = K c (RT) n 3.2x10-4 = K c [(0.0821)(300 K)] 1 = 1.30x10-5 What is the value for K c for: NO 2 (g) ½ N 2 O 4 (g) K c = (1.30x10-5 ) -½ = 277

3 5. Given the following equations: H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Kc = 4.8 FeO (s) + CO (g) Fe (s) + CO 2 (g) K c = 0.48 Calculate the K c value for: Fe (s) + H 2 O (g) FeO (s) + H 2 (g) K c =??? H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Kc = 4.8 Fe (s) + CO 2 (g) FeO (s) + CO (g) K c = (0.48) -1 = 2.08 Fe (s) + H 2 O (g) FeO (s) + H 2 (g) K c = (4.8)(2.08) = Given the following equations: S (s) + O 2 (g) SO 2 (g) K p = SO 3 (g) 2 SO 2 (g) + O 2 (g) K p = Calculate the K p value for: S (s) + 3/2 O 2 (g) SO 3 (g) K p =??? S (s) + O 2 (g) SO 2 (g) K p = 48.2 SO 2 (g) + ½ O 2 (g) SO 3 (g) K p = (0.075) -1/2 = 3.65 S (s) + 3/2 O 2 (g) SO 3 (g) K p = Which of the following equations has K c = K p a. PCl 5 (g) PCl 3 (g) + Cl 2 (g) No; gas = +1 b. 2 NOCl (g) 2 NO (g) + Cl 2 (g) No; gas = +1 c. CaCO 3 (s) CaO (s) + CO 2 (g) No; gas = +1 d. H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Yes; gas = 0 e. 2 NO (g) N 2 (g) + O 2 (g) Yes; gas = 0 K c and K p Calculations At Equilibrium 1. For the reaction: 2 NO 2 (g) N 2 O 4 (g) At equilibrium [N 2 O 4 ] = 0.25 M & [NO 2 ] = M. Calculate K c K c = [N 2 O 4 ] K c = [0.25] K c = 8.16 [NO 2 ] 2 [0.175] 2 2. For the reaction: 2 NH 3 (g) N 2 (g) + 3 H 2 (g) K p = 32 At equilibrium P NH3 = 0.64 atm & P N2 = 1.18 atm. Calculate P H2 K p = P N2 P 3 H2 P 2 NH3 32 = (1.18)P 3 H2 (0.64) 2 P H2 = 2.23

4 Dissociation 3. For the reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) If the initial pressure of PCl 5 is 2 atm and at equilibrium it is 15% dissociated, what is K p? PCl 5 (g) PCl 3 (g) + Cl 2 (g) P o P eq For the reaction: 2 NO (g) N 2 (g) + O 2 (g) If the initial [NO] = 0.50 M and at equilibrium it is 5% dissociated, what is K c? 2 NO (g) N 2 (g) + O 2 (g) [] o [] eq K p = P PCl3 P Cl2 P PCl5 K p = [0.30] 2 K p = K c = [N 2 ] [O 2 ] [NO] 2 K c = [0.0125] 2 K c = 6.9x10-4 [0.475] 2 5. For the equation: NH 4 I (s) NH 3 (g) + HI (g) The total pressure at equilibrium is 4.2 atm. What is K p? If P tot = 4.2 atm 2x = 4.2 x = 2.1 atm K p = P NH3 P HI = (2.1) 2 = For the equation: (NH 4 )(H 2 NCO 2 ) (s) 2 NH 3 (g) + CO 2 (g) The total pressure at equilibrium is 0.33 atm. What is K p? If P tot = 0.33 atm 3x = 0.33 x = 0.11 K p = P 2 NH3P CO2 = (0.22) 2 (0.11) = 5.32x For the equation: N 2 (g) + O 2 (g) 2 NO (g), you start with 2 M of each of the reactants. They react away to an extent of 27% to reach equilibrium. Calculate the value of K c. N 2 (g) + O 2 (g) 2 NO (g) [] o [] eq K c = [NO] 2 [N 2 ] [O 2 ] K c = [1.08] 2 K c = [1.46] [1.46] 8. For the equation: 2 NOBr (g) 2 NO (g) + Br 2 (g), you start with 0.75 M of the NOBr. At equilibrium, the NOBr has reacted away by 89%. Calculate the value of K c. 2 NOBr (g) 2 NO (g) + Br 2 (g) [] o [] eq K c = [NO] 2 [Br 2 ] [NOBr] 2 K c = [0.668] 2 [0.334] K c = 22.2 [0.082] 2

5 Q vs. K 9. For the reaction: 2 NOCl (g) 2 NO (g) + Cl 2 (g) K c = 1.2x10-3 If the initial [NOCl] o = 0.15 M, [NO] o = 0.75 M, and [Cl 2 ] o = 0.05 M, is the system at equilibrium? If not, which way will the reaction shift, left or right? K c = [NO] 2 [Cl 2 ] [NOCl] 2 Q = [0.75] 2 [0.05] Q = 1.25 > 1.2x10-3 so the reaction goes left [0.15] For the reaction: NH 3 (aq) + H 2 O (l) NH 4 +1 (aq) + OH -1 (aq) K c = 1.8x10-5 If the initial [NH 3 ] o = 0.5 M, [NH 4 +1 ] = M, and [OH -1 ] = M, is the system at equilibrium? If not, which way will the reaction shift, left or right? K c = [NH +1 4 ][OH -1 ] [NH 3 ] Q = [0.0025] 2 Q = 1.25x10-5 < 1.8x10-5 so the reaction goes right [0.5] 11. For the equation: CS 2 (g) + 3 Cl 2 (g) S 2 Cl 2 (g) + CCl 4 (g), K c = 4.8x10-2. If you start with [CS 2 ] = M, [Cl 2 ] = M, [S 2 Cl 2 ]= 0.58 M, and [CCl 4 ] = M, is the reaction at equilibrium? If not, which way will the reaction go to reach equilibrium (left or right)? Q Q = = [S [S s Cl 2 Cl 2 2 ][CCl 4 ] 4 ] Q Q = = (0.58)(0.042) = 2.3 CS 2 (g) + 3 Cl 2 (g) S 2 Cl 2 (g) + CCl 4 [CS [CS 2 ][Cl 2 ] ][Cl 2 ] 3 (0.025)(0.175) 3 [] o Q Q = > 182 K >>>> 4.8 x10-2 c so reaction goes left so rxn goes left Calculating Equilibrium Conditions 12. For the equation: PCl 5 (g) PCl 3 (g) + Cl 2 (g), you start with 0.25 atm of each of the products as well as the reactants. The K p value is Is the reaction at equilibrium? Prove it. What are the equilibrium pressures of all species? PCl 5 (s) PCl 3 (g) + Cl 2 (g) P o x -x -x P eq x x x Q = P PCl3 P Cl2 Q = (0.25) 2 = 0.25 P PCl5 (0.25) Q > K p 0.25 > so there are too many products so the reaction moves to the left K p = P PCl3 P Cl = (0.25-x) 2 x = P PCl5 (0.25+x) P PCl5 = P PCl3 = P Cl2 = For the equation: H 2 O (g) + CO (g) CO 2 (g) + H 2 (g) K c = If 2 moles of each of H 2 O and CO are put into a 10 L container, what is the concentration of all species at equilibrium? H 2 O (g) + CO (g) CO 2 (g) + H 2 (g) P o x -x +x +x P eq x 0.2 -x x x K c = [CO 2 ][H 2 ] = (x) 2 [H 2 O][CO] (0.2-x) 2 x = [CO 2 ] = [H 2 ] = [H 2 O] = [CO] = 0.135

6 14. For the equation: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K p = 4.8 If enough SO 2 Cl 2 is put into a container so its pressure is 8 atm, what is the equilibrium pressure of all species. What is the total pressure? SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) P o x +x +x P eq 8-x x x K p = P SO2 P Cl2 P SO2Cl2 K p = [x] 2 K p = x P SO2 = P Cl2 = 4.25 P SO2Cl2 = 3.75 P tot = = For the equation: CaCO 3 (s) CaO (s) + CO 2 (g) K p = 2.4 If 200 g CaCO 3 is put into a 20 L container at 500 K, how many grams of it remain at equilibrium? CaCO 3 (s) CaO (s) + CO 2 (g) P o 0 +x P eq x K p = 2.4 = P CO2 P CO2 = 2.4 PV = nrt (2.4 atm)(20 L) = n( Latm/molK)(500 K) n = 1.17moles reacted away 1.17 moles CaCO 3 * (100.1 g/mole) = 117 g reacted 200 g to begin with 117 g reacted = 83 g left 16. For the equation: 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g), you start with some KClO 3 that decomposes into the products. At equilibrium, there is some solid remaining and the total pressure in the flask is 0.29 atm. Calculate the value of K p. 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) P o P eq 0.29 K P = P 3 O2 =

7 17. For the equation: NH 4 Cl (s) NH 3 (g) + HCl (g), you start with some NH 4 Cl that decomposes into the products. At equilibrium, there is some solid remaining and the total pressure in the flask is 1.8 atm. Calculate the value of K p. NH 4 Cl (s) NH 3 (g) + HCl (g) P o P eq K P = P NH3 P HCl = (0.9)(0.9) = For the equation: COBr 2 (g) CO (g) + Br 2 (g), you start with 4 moles in a 10 L vessel of COBr 2. The reaction has a K c = What are the equilibrium concentrations of all species? COBr 2 (g) CO (g) + Br 2 (g) [] o x +x +x [] eq 0.4-x x x 19. For the equation: H 2 (g) + CO 2 (g) H 2 O (g) + CO (g), you start with 2 atm of each of the reactants and none of the products. The K p = 3.4. What are the equilibrium pressures of all species? H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) P o x -x +x +x P eq 2-x 2-x x x K c = [CO][Br 2 ] [COBr 2 ] 0.76 = x 2 x = 0.29 [0.4-x] [COBr 2 ] = 0.11 [CO] = [Br 2 ] = 0.29 K p = P H2O P CO P H2 P CO2 3.4 = x 2 x = 1.3 [2-x] 2 P CO2 = P H2 = 0.7 P CO = P H2O = For the equation: 2 CH 2 Cl 2 (g) CH 4 (g) + CCl 4 (g), you start with 0.25 M of CH 2 Cl 2 and is has a K c value of What are the equilibrium concentrations of all species? 2 CH 2 Cl 2 (g) CH 4 (g) + CCl 4 (g) K c = [CH 4 ][CCl 4 ] [] o [CH 2 Cl 2 ] 2-2x +x +x 0.84 = x 2 x = [] eq x x x [0.25-2x] 2 [CH 2 Cl 2 ] = [CH 4 ] = [CCl 4 ] = For the equation: NH 4 HS (s) NH 3 (g) + H 2 S (g), you start with 100 grams of NH 4 HS (s) in a 2.5 L flask at 500 K. The K p value is How many grams of the solid remain at equilibrium? NH 4 HS (s) NH 3 (g) + H 2 S (g) P o 0 0 +x +x P eq x x K P = P NH3 P H2S x 2 = 1.45 x = 1.20 atm n = PV/RT = (1.2 atm)(2.5 L)/[(0.0821)(500 K)] n = moles NH 3 so there must have been the same moles of NH 4 HS that broke up ( mol)(51.1 g/mol) = g broke up so: 100 g g = g remain

8 LeChatelier s Principle 1. State LeChatelier s Principle.A change in any factor that effects the equilibrium of a system will cause the system to shift in such a way to reduce or counteract that initial change. 2. For the following reaction: Heat + CaCO 3 (s) + CO 2 (g) + H 2 O (l) Ca +2 (aq) + 2 HCO -1 3 (aq) What will be the effect of doing each of the following actions on the above equilibrium? a) Adding CaCO 3 (s) Left Right No Change b) Removing Ca +2 (aq) Left Right No Change c) Removing CO 2 (g) Left Right No Change d) Adding NaHCO 3 (s) Left Right No Change e) Adding Ne (g) Left Right No Change f) Adding CO 2 (g) Left Right No Change g) Increasing temperature Left Right No Change h) Decreasing volume Left Right No Change 3. For the following reaction: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) + Heat What will be the effect of doing each of the following actions on the above equilibrium? a) Decreasing temperature Left Right No Change b) Increasing O 2 (g) Left Right No Change c) Decreasing SO 2 (g) Left Right No Change d) Increasing volume Left Right No Change e) Increasing SO 3 (g) Left Right No Change f) Adding N 2 (g) Left Right No Change

9 K sp 1. For each of the substances below, write the solubility equation as well as the K sp equation. a. AgCl AgCl Ag + + Cl -1 K sp = [Ag + ][Cl -1 ] b. PbI 2 PbI 2 Pb I -1 K sp = [Pb +2 ][I -1 ] 2 c. Ag 2 CO 3 Ag 2 CO 3 2 Ag CO 3 d. Ca 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2 3 Ca PO 4 K sp = [Ag + ] 2 [CO -2 3 ] K sp = [Ca +2 ] 3 [PO -3 4 ] 2 Problem Solving with K sp 1. Calculate the K sp of CaCrO 4 if a saturated solution has [Ca +2 ] = 4.5 x10-5 CaCrO 4 Ca CrO 4 K sp = [Ca +2 ][CrO -2 4 ] = (4.5 x10-5 ) (4.5 x10-5 ) = x Calculate the K sp of Fe(OH) 3 if a saturated solution has [Fe +3 ] = 4.2 x10-6 Fe(OH) 3 Fe OH -1 K sp = [Fe +3 ][OH -1 ] 3 = (4.2 x10-6 ) [(3)(4.2 x10-6 )] 3 = 8.4 x Calculate the solubility (moles/l) of PbCO 3 if K sp = 7.4 x10-14 PbCO 3 Pb CO 3 K sp = [Pb +2 ][CO -2 3 ] 7.4x10-14 = x 2 x = 2.72 x10-7 mole/l 4. Calculate the solubility (moles/l) of Ag 2 SO 4 if K sp = 1.2 x10-5 Ag 2 SO 4 2 Ag SO 4 K sp = [Ag +1 ] 2 [SO -2 4 ] 1.2x10-5 = 4x 3 x = mole/l 5. Calculate the solubility (mg/l) of FePO 4 if K sp = 9.4 x10-9 FePO 4 Fe PO 4 K sp = [Fe +3 ][PO -3 4 ] 9.4x10-9 = x 2 x = 9.70 x10-5 mole/l 6. Calculate the solubility (mg/l) of Al 2 (CO 3 ) 3 if K sp = 7.2 x10-25 Al 2 (CO 3 ) 3 2 Al CO 3 K sp = [Al +3 ] 2 [CO -2 3 ] 3 7.2x10-25 = 108x 5 x = 5.82 x10-6 mole/l *(234 g/mole) *(1000 mg/g) = 1.36 mg/l 7. How many mg of CuCrO 4 will dissolve in 50 ml of water (K sp = 9.4 x10-10 ) CuCrO 4 Cu CrO 4 K sp = [Cu +2 ] [CrO -2 4 ] 9.4x10-10 = x 2 x = 3.07 x10-5 mole/l *(179.5 g/mole) *(1000 mg/g)*(0.05 L) = mg 8. If 100 mg of CaCO 3 (K sp = 3.4 x10-9 ) is put in 500 ml of water, how many mg remain undissolved? CaCO 3 Ca CO 3 K sp = [Ca +2 ] [CO -2 3 ] 3.4x10-9 = x 2 x = 5.83 x10-5 mole/l *(179.5 g/mole) *(1000 mg/g)*(0.5 L) = 2.92 mg 100 mg 2.92 mg dissolved = 97.1 mg undissolved 9. Put the following substances in order of least soluble to most soluble. NiCO 3 (K sp = 1.4x10-7 ) AuCl (K sp = 2x10-13 ) MnCO 3 (K sp = 2.3x10-11 ) PbCrO 4 (K sp = 2.8x10-13 ) Since they are all the same number of pieces, you can compare them directly by K sp values. Thus: AuCl (K sp = 2x10-13 ) < PbCrO 4 (K sp = 2.8x10-13 ) < MnCO 3 (K sp = 2.3x10-11 ) < NiCO 3 (K sp = 1.4x10-7 ) 10. Put the following substances in order of least soluble to most soluble. Zn(CN) 2 (K sp = 8x10-12 ) AgBr (K sp = 5.x10-13 ) Pb(OH) 2 (K sp = 1.4x10-15 ) BaSO 4 (K sp = 1.1x10-10 ) These substances have different number of ionic pieces so you can t compare K sp values directly. If you do a solubility ICE table for each, the ultimate results will be as follows where x is the solubility: Zn(CN) 2 4x 3 = 8x10-12 x = 1.27x10-4 M AgBr x 2 = 5x10-13 x = 7.1x10-7 M Pb(OH) 2 4x 3 = 1.4x10-15 x = 7.1x10-6 M BaSO 4 x 2 = 1.1x10-10 x = 1.1x10-5 M Thus AgBr < Pb(OH) 2 < BaSO 4 < Zn(CN) 2

10 11. Prove which of each of the substances below is the most soluble. a. AgBr (K sp = 5.4x10-13 ) vs. AgI (K sp = 8.5 x10-17 ) AgBr Ag +1 + Br -1 K sp = [Ag +1 ][Br -1 ] 5.4x10-13 = x 2 x = 7.3 x10-7 mole/l AgI Ag +1 + I -1 K sp = [Ag +1 ][I -1 ] 8.5x10-17 = x 2 x = 9.2 x10-9 mole/l b. PbCl 2 (K sp = 1.7 x10-5 ) vs. PbBr 2 (K sp = 6.6 x10-6 ) PbCl 2 Pb Cl -1 K sp = [Pb +2 ][Cl -1 ] 2 1.7x10-5 = 4x 3 x = mole/l PbBr 2 Pb Br -1 K sp = [Pb +2 ][Br -1 ] 2 6.6x10-6 = 4x 3 x = mole/l c. AgCl (K sp = 1.8 x10-10 ) vs. Ag 2 CrO 4 (K sp = 1.1 x10-12 ) AgCl Ag +1 + Cl -1 K sp = [Ag +1 ][Cl -1 ] 1.8x10-10 = x 2 x = 1.3 x10-5 mole/l Ag 2 CrO 4 2 Ag CrO 4 K sp = [Ag +1 ] 2 [CrO -2 4 ] 1.1x10-12 = 4x 3 x = 6.5x10-5 mole/l d. CaCO 3 (K sp = 3.4 x10-9 ) vs. Mg(OH) 2 (K sp = 5.6 x10-12 ) CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO -2 3 ] 3.4x10-9 = x 2 x = 5.8 x10-5 mole/l Mg(OH) 2 Mg OH -1 K sp = [Mg +2 ][OH -1 ] 2 5.6x10-12 = 4x 3 x = 1.12 x10-4 mole/l Common Ion Effect 1. Calculate the solubility (moles/l) of MgS (K sp = 5.2 x10-16 ) in: a. Pure water MgS Mg +2 + S -2 K sp = [Mg +2 ][S -2 ] 5.2x10-16 = x 2 x = 2.3 x10-8 mole/l b. A 0.25 M solution of MgCl 2 MgS Mg +2 + S -2 K sp = [Mg +2 ][S -2 ] 5.2x10-16 = (0.25)(x) x = 2.1 x10-15 mole/l 2. Calculate the solubility (moles/l) of PbCl 2 (K sp = 1.7 x10-5 ) in: a. Pure water PbCl 2 Pb Cl -1 K sp = [Pb +2 ][Cl -1 ] 2 1.7x10-5 = 4x 3 x = mole/l b. A 0.55 M solution of NaCl PbCl 2 Pb Cl -1 K sp = [Pb +2 ][Cl -1 ] 2 1.7x10-5 = (x)(0.55) 2 x = 5.6x10-5 mole/l 3. Calculate the solubility (moles/l) of PbI 2 (K sp = 9.8 x10-9 ) in: a. Pure water PbI 2 Pb I -1 K sp = [Pb +2 ][I -1 ] 2 9.8x10-9 = 4x 3 x = 1.35x10-3 mole/l b. A M solution of AlI 3 PbI 2 Pb I -1 K sp = [Pb +2 ][I -1 ] 2 9.8x10-9 = (x)(0.015) 2 x = mole/l 4. Calculate what mass of Hg 2 SO 4 (K sp = 6.5 x10-7 ) will dissolve per liter in: a. Pure water Hg 2 SO 4 2 Hg SO 4 K sp = [Hg +1 ] 2 [SO -2 4 ] 6.5x10-7 = 4x 3 x = 5.46x10-3 mole/l(497.3 g/mol) = 2.71 g/l b. A M solution of Na 2 SO 4 Hg 2 SO 4 2 Hg SO 4 K sp = [Hg +1 ] 2 [SO -2 4 ] 6.5x10-7 = (4x 2 )(0.0075) x = 4.65x10-3 mole/l (497.3 g/mol) = 2.31 g/l

11 5. Calculate what mass of CaCO 3 (K sp = 3.4 x10-9 ) will dissolve per liter in: a. Pure water CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO -2 3 ] 3.4x10-9 = x 2 x = 5.83 x10-5 mole/l (100.1 g/mole) = 5.84 x10-3 g/l b. A 0.45 M solution of CaCl 2 CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO -2 3 ] 3.4x10-9 = (0.45)(x) x = 7.56 x10-9 mole/l (100.1 g/mole) = 7.56 x10-7 g/l 6. Calculate what mass of PbCl 2 (K sp = 1.7 x10-5 ) will dissolve per liter in: a. Pure water PbCl 2 Pb Cl -1 K sp = [Pb +2 ][Cl -1 ] 2 1.7x10-5 = 4x 3 x = mole/l(278.2 g/mole) = 4.5 g/l b. A M solution of AlCl 3 PbCl 2 Pb Cl -1 K sp = [Pb +2 ][Cl -1 ] 2 1.7x10-5 = (x)(0.201) 2 x = 4.2x10-4 mole/l(278.2 g/mole) = g/l 7. AgBr will be the least soluble in 0.10 M: a. NaBr b. CaBr 2 c. AlBr 3 d. Ag 2 SO 4 e. AgNO 3 AlBr 3 will have the most concentrated [Br -1 ] or [Ag +1 ] since there are 3 Br -1 for every 1 AlBr 3. Thus, AgBr will be least soluble in it. 8. MgF 2 will be least soluble in 0.25 M: a. KF b. Mg(NO 3 ) 2 c. MgC 2 O 4 d. BaF 2 e. LiF BaF 2 will have the most concentrated [F -1 ] or [Mg +2 ] since there are 2 F -1 for every 1 BaF 2. Thus, MgF 2 will be least soluble in it. Precipitation 1. Will a ppt of CaCO 3 (K sp = 3.4 x10-9 ) form if [Ca +2 ] = 4 x10-6 M and [CO 3-2 ] = 4 x10-3? Q = [4 x10-6 ][4x10-3 ] = 1.6x10-8 >> 3.4x10-9 so yes, ppt 2. Will a ppt of Ag 2 CrO 4 (K sp = 1.1 x10-12 ) form if [Ag + ] = 3x10-4 and [CrO 4-2 ] = 2x10-4? Q = [3 x10-4 ] 2 [2x10-4 ] = 1.8x10-11 >> 1.1x10-12 so yes, ppt 3. Will a ppt of BaCO 3 (K sp = 2.6 x10-9 ) form if 50 ml of 4x10-5 M Ba +2 is mixed with 50 ml of 8x10-5 M CO 3-2? Q = [(50/100)4 x10-5 ][(50/100)8x10-5 ] = 8x10-10 << 2.6x10-9 so no ppt 4. Will a ppt of PbBr 2 (K sp = 6.6 x10-6 ) form if 150 ml of 2x10-2 M Pb(NO 3 ) 2 is mixed with 50 ml of 3x10-2 M AlBr 3? Q = [(150/200)2 x10-2 ][(50/200)(3) 3x10-2 ] = 7.6x10-6 >> 6.6x10-6 so yes, ppt 5. What concentration of [OH -1 ] will cause a ppt of Fe(OH) 2 (K sp = 4.9x10-17 ) of a 2x10-5 M Fe +2 solution? 4.9x10-17 = (2x10-5 )(OH -1 ) 2 [OH -1 ] = 1.56x10-6 M 6. What concentration of [C 2 O 4-2 ] will cause a ppt of Ag 2 C 2 O 4 (K sp = 5.4x10-12 ) of a 4x10-4 M Ag +1 solution? 5.4x10-12 = (4x10-4 ) 2 (C 2 O 4-2 ) [C 2 O 4-2 ] = 3.4x10-5 M 7. What mass of Ni(NO 3 ) 2 *7 H 2 O (s) will cause a ppt of NiCO 3 (K sp = 1.4x10-7 ) in a 2 L solution of 3.5x10-4 M CO 3-2? 1.4x10-7 = [Ni +2 ](3.5x10-4 ) [Ni +2 ] = 4x10-4 M Molar mass of Ni(NO 3 ) 2 *7H 2 O = g/mole [Ni +2 ] = 4x10-4 moles/l * 2 L * g/mole = grams Ni(NO 3 ) 2 *7H 2 O 8. What mass of Pb(NO 3 ) 2 will cause a ppt of PbCl 2 (K sp = 1.7x10-5 ) in 100 ml of a 2.4x10-2 M solution of AlCl 3? Since there are 3 Cl -1 ions for every AlCl 3 then [Cl -1 ] = 7.2x10-2 M so: 1.7x10-5 = [Pb +2 ](0.072) 2 [Pb +2 ] = 3.28x10-3 M Molar mass of Pb(NO 3 ) 2 is g/mole Thus: 3.28x10-3 moles/l * 1 L/1000 ml * 100 ml * g/mole = grams Pb(NO 3 ) 2

12 Separations by K sp 1. A solution is made so that [Ca +2 ] = and [Mg +2 ] = as well. If Cr 2 O 7-2 is added, answer the following questions. Know that K sp CaCr 2 O 7 = 4.8 x10-7 and K sp MgCr 2 O 7 = 7.6 x10-8 a. Which will ppt 1 st? At what [Cr 2 O -2 7 ] will it begin to ppt? CaCr 2 O 7 Ca Cr 2 O 7 K sp = [Ca +2 ][ Cr 2 O -2 7 ] 4.8x10-7 = x(0.0005) x = 9.6 x10-4 mole/l MgCr 2 O 7 Mg Cr 2 O 7 K sp = [Mg +2 ][ Cr 2 O -2 7 ] 7.6x10-8 = x(0.0005) x = 1.5 x10-4 mole/l MgCr 2 O 7 ppts 1 st b. What is the maximum [Cr 2 O -2 7 ] that can be made to ppt almost all of one and none of the other? 9.6x10-4 moles/l c. What is the concentration of the less soluble ion under the conditions specified in b? MgCr 2 O 7 Mg Cr 2 O 7 K sp = [Mg +2 ][ Cr 2 O -2 7 ] 7.6x10-8 = (9.6x10-4 )x x = 7.92 x10-5 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (7.92x10-5 /0.0005)*100% = 15.8% 2. A solution is made so that [Pb +2 ] = and [Hg +2 ] = If SO -2 4 is added, answer the following questions. Know that K sp PbSO 4 = 2.5 x10-8 and K sp HgSO 4 = 5.4 x10-7 a. Which will ppt 1 st? At what [SO -2 4 ] will it begin to ppt? PbSO 4 Pb SO 4 K sp = [Pb +2 ][ SO -2 4 ] 2.5x10-8 = x(0.0025) x = 1 x10-5 mole/l HgSO 4 Hg SO 4 K sp = [Hg +2 ][ SO -2 4 ] 5.4x10-7 = x(0.0075) x = 7.2 x10-5 mole/l PbSO 4 ppts 1 st b. What is the maximum [SO -2 4 ] that can be made to ppt almost all of one and none of the other? HgSO 4 Hg SO 4 K sp = [Hg +2 ][ SO -2 4 ] 5.4x10-7 = x(0.0075) x = 7.2 x10-5 mole/l c. What is the concentration of the less soluble ion under the conditions specified in b? PbSO 4 Pb SO 4 K sp = [Pb +2 ][ SO -2 4 ] 2.5x10-8 = x(7.2x10-5 ) x = 3.5 x10-4 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (3.5x10-4 /0.0025)*100% = 13.9% 3. A solution is made so that [Zn +2 ] = and [Ag +1 ] = If F -1 is added, answer the following questions. Know that K sp ZnF 2 = 4.8 x10-7 and K sp AgF = 7.6 x10-8 a. Which will ppt 1 st? At what [F -1 ] will it begin to ppt? ZnF 2 Zn F -1 K sp = [Zn +2 ][ F -1 ] 2 4.8x10-7 = x 2 (0.0025) x = 1.39 x10-2 mole/l AgF Ag +1 + F -1 K sp = [Ag +1 ][ F -1 ] 7.6x10-8 = x(0.057) x = 1.3 x10-6 mole/l AgF ppts 1 st b. What is the maximum [F -1 ] that can be made to ppt almost all of one and none of the other? 1.39x10-2 mole/l c. What is the concentration of the less soluble ion under the conditions specified in b? AgF Ag +1 + F -1 K sp = [Ag +1 ][ F -1 ] 7.6x10-8 = 1.39x10-2 (x) x = 5.47 x10-6 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (5.47x10-6 /0.057)*100% = %

13 Combination Problems 1. For the reaction at 2000 K H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) For an experiment, the equilibrium values of each substance are as follows: [H 2 ] = 0.20 M [CO 2 ] = 0.30 M [H 2 O] = [CO] = 0.55 M a. What is the mole fraction of CO in the equilibrium mixture? X CO = (0.55)/( ) = b. Calculate the value of K c, the equilibrium constant for the reaction above. K c = (0.55) 2 /(0.20)(0.30) = c. Determine K p in terms of K c for this system. K p = K c because n gas = 0 d. When the system is cooled from 2000 K to a lower temperature, 30% of the CO is converted back to CO 2. Calculate the value of K c at this lower temperature. Since 30% of CO is converted: 0.55*0.30 = changes for each H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) [] [] eq K c = (0.385) 2 /(0.365)(0.465) = e. In a different experiment, 0.50 mole of H 2 is mixed with 0.50 mole of CO 2 in a 3.0 L reaction vessel at 2000 K. Calculate the equilibrium concentration, in M, of CO at this temperature. H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) [] x -x +x +x [] eq x x x x = x 2 /(0.167-x) = x/(0.167-x) x = [CO] =

14 2. For the reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) It is observed that greater amounts of PCl 3 and Cl 2 are produced as the temperature is increased. a) What is the sign of S o for the reaction? Explain. S should be positive. You are going from one mole of gas on the left to 2 moles on the rights. This shows an increase in disorder and a positive entropy change b) What change, if any, will occur in G o for the reaction as the temperature is increased? Explain. Since S is positive, G should decrease as temperature is increased. This is due to the equation G = H T S. Since T is in K and therefore positive, you will be subtracting a positive value from G. c) If He gas is added to the original mixture at constant volume and temperature, what will happen to the partial pressure of Cl 2? Explain. Dalton s law of partial pressures says that the total pressure is equal to the sum of the individual pressures. Adding more He gas will increase the total pressure but not change the pressure the Cl 2 exerts. d) If the volume of the reaction mixture is decreased at constant temperature to half the original volume, what will happen to the number of moles of Cl 2 in the reaction vessel? Explain. By decreasing the volume, you are increasing the pressures of each of the above species. Since there are unequal amounts of gas on the reactant and product sides of the equation above, the system will try to reduce the pressure increase it experienced. It can do this by going to the left side which will reduce the number of moles of Cl 2

15 3. For the reaction: C(s) + CO 2 (g) 2 CO(g) Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are recorded in the table below. Time (hours) Total Pressure of gases (atm) (a) Write the expression for the equilibrium constant, Kp, for the reaction. K p = P 2 CO/P CO2 (b) Calculate the number of moles of CO 2 (g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.) n = PV/RT = (5 atm)(2 L)/(0.0821)(1160 K) = moles (c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO 2 (g) is 1.63 atm. Calculate: (i) the partial pressure of CO(g), and CO 2 : 5 atm 1.63 atm = 3.37 atm CO 2 reacted *(2 mole CO/1 mole CO 2 )= 6.74 atm CO (ii) the value of the equilibrium constant, Kp. K p = P 2 CO/P CO2 = (6.74) 2 /1.63 = (d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.) Nothing would change. It would just get to equilibrium faster. (e) In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s), plus CO(g) and CO 2 (g), each at a partial pressure of 2.00 atm at 1,160 K. Predict whether the partial pressure of CO 2 (g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation. Q = (2) 2 /2 = 2 Q <<< K p so reaction will go to the right This will cause the partial pressure of CO 2 to decrease

16 4. Answer the following questions: a. A saturated solution is prepared by adding excess PbI 2 (s) to distilled water to form 1.0 L of solution at 25 o C. The concentration of Pb +2 (aq) in the saturated solution is found to be 1.3x10-3 M. The chemical equation for the dissolution of PbI 2 (s) in water is shown below: PbI 2 (s) Pb +2 (aq) + 2 I -1 (aq) i) Write the equilibrium-constant expression for the equation. K sp = [Pb +2 ][I -1 ] 2 ii) Calculate the molar concentration of I -1 (aq) in the solution. [Pb +2 ] = 1.3x10-3 M *(2 moles I -1 /1 mole Pb +2 ) = 2.6x10-3 M iii) Calculate the value of the equilibrium constant, K sp. K sp = (1.3x10-3 )(2.6x10-3 ) 2 = 8.79x10-9 b. A saturated solution is prepared by PbI 2 (s) to distilled water to form 2.0 L of solution at 25 o C. What are the molar concentrations of Pb +2 (aq) and I -1 (aq) in the solution? Justify your answer. The concentrations should be the same as above; [Pb +2 ] = 1.3x10-3 and [I -1 ] = 2.6x10-3. Changing the volume does not change the concentration if the solution is still saturated. c. Soild NaI is added to a saturated solution of PbI 2 at 25 o C. Assuming that the volume of the solution does not change, does the molar concentration of Pb +2 (aq) in the solution increase, decrease, or stay the same? Justify your answer. By adding NaI, you are adding [I -1 ] and thus, by LeChatelier s Principle, the reaction should go to the left and form a precipitate. To do this, you must use up some of the Pb +2 so thus [Pb +2 ] should decrease 5. The value of K sp for the salt BaCrO 4 is 1.2x When a 500. ml sample of 8.2x10-6 M Ba(NO 3 ) 2 is added to 500. ml of 8.2x10-6 M Na 2 CrO 4, no precipitate is observed. i) Assuming the volumes are additive, calculate the molar concentrations of Ba +2 (aq) and CrO 4-2 (aq) in the 1.00 L of solution. [Ba +2 ] = 8.2x10-6 M Ba(NO 3 ) 2 *(500/1000) = 4.1x10-6 M [CrO 4-2 ] = 8.2x10-6 M Na 2 CrO 4 *(500/1000) = 4.1x10-6 M ii) Use the molar concentrations of Ba +2 (aq) and CrO -2 4 (aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer. Q = [Ba +2 ][CrO -2 4 ] = (4.1x10-6 )(4.1x10-6 ) = 1.681x10-11 Q < K sp 1.681x10-1 < 1.2x10-10 so no ppt forms

17 6. Silver chromate dissociates in water according the equation below: Ag 2 CrO 4 (s) 2 Ag +1 (aq) + CrO 4-2 (aq) K sp = 2.6x10-12 at 25 o C a) Write the equilibrium-constant expression for the dissolving of Ag 2 CrO 4 (s) K sp = [Ag +1 ] 2 [CrO 4-2 ] b) Calculate the concentration, in M, of Ag +1 (aq) in a saturated solution of Ag 2 CrO 4 at 25 o C 2.6x10-12 = 4x 3 x = 8.66x10-5 M Since there are 2 Ag +1 ions present, [Ag +1 ] = 1.732x10-4 c) Calculate the maximum mass, in grams, of Ag 2 CrO 4 that can dissolve in 100. ml of water at 25 o C. x = 8.6x10-5 moles/l *(0.10 L) *(331.8 g/mole) = g d) A mol sample of solid AgNO 3 is added to a 1.00 L saturated solution of Ag 2 CrO 4. Assuming no volume change, does [CrO 4-2 ] increase, decrease, or stay the same? Justify. Since AgNO 3 will dissolve and make more Ag +1 ions, the reaction will go left due to Le Chatelier s principle. This will cause CrO 4-2 to react and ppt so [CrO 4-2 ] will decrease. 7. In a saturated solution of Ag 3 PO 4 at 25 o C, the concentration of Ag +1 (aq) is 5.3x10-5 M. The equilibrium constant expression for the dissolving of Ag 3 PO 4 (s) in water is shown below: K sp = [Ag +1 ] 3 [PO 4-3 ] a) Write the balanced equation for the dissolving of Ag 3 PO 4 in water. Ag 3 PO 4 3 Ag +1 + PO 4-3 b) Calculate the value of K sp for Ag 3 PO 4 at 25 o C. [Ag +1 ] = 5.3x10-5 but there is only 1 PO 4-3 for every 3 Ag +1 so: 5.3x10-5 *(1 mole PO 4-3 /3 mole Ag +1 ) = 1.77 x10-5 M PO 4-3 K sp = [Ag +1 ] 3 [PO 4-3 ] = (5.3x10-5 ) 3 (1.77x10-5 ) = 2.635x10-18 c) A 1.00 L sample of saturated Ag 3 PO 4 solution is allowed to evaporate at 25 o C to a final volume of 500. ml. What is the [Ag +1 ] in the solution? Justify your answer. The concentration does not change in a saturated solution. As 500 ml evaporates, solid Ag 3 PO 4 will ppt out at the bottom of the beaker. The solution will still be saturated, though, and therefore not change its concentration.

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