! b. Calculate the ph of the saturated solution. (Hint: How many OH ions form for every Zn(OH) 2 that dissolves? Calculate poh, then ph.)! (8.

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1 AP Chem Worksheet: Solubility Product, K sp Page 1 Write your chemical equations for dissolving the solid and the K sp expression before trying to solve the problems!! 1. The molar solubility of copper(i) chloride, CuCl, is 1.1 x 10 3 M at 25 C. Calculate K sp. (1.2 x10 6 ) 2. The molar solubility of Barium iodate, Ba(IO 3 ) 2, at 35 C is 1.0 x 10 3 M. Calculate K sp at this temperature.! (4 x10 9 ) 3. a. Look up K sp and calculate the molar solubility ( how much dissolves ) of Zn(OH) 2 in plain water.! (2.0 x10 6 M)! b. Calculate the ph of the saturated solution. (Hint: How many OH ions form for every Zn(OH) 2 that dissolves? Calculate poh, then ph.)! (8.60)! c. How many grams of Zn(OH) 2 dissolve per liter of solution?! (2.0 x10 4 g/l) 4. a. Look up K sp and calculate the molar solubility ( how much dissolves ) of AgI in plain water.! (9.1 x10 9 )! b. Calculate the molar solubility of AgI in M NaI solution. Why is it less soluble than in plain water?! (1.6x10 15 ) 5. a. Calculate the molar solubility of Cd(OH) 2 in plain water at 25 C.! (1.9x10 5 )! b. Calculate the ph of the saturated solution.! (9.58)

2 5c. Calculate the molar solubility of Cd(OH) 2 in a buffer solution with a ph of 9.00! Page 2! (Hint: Calculate the [OH ], from outside, in the buffer.)! (2.5x10 4 M) d. Calculate the molar solubility of Cd(OH) 2 in a buffer solution with a ph of 7.00 (2.5 M) e. Why does the Cd(OH) 2 have different solubilites depending on the ph? Calculating the concentration of Ions in a solution 6. Some KMnO 4 is dissolved in water. Ba(NO 3 ) 2 is added so Ba(MnO 4 ) 2 ppts. The equilibrium [Ba +2 ]= 2.0 x 10 8 M. Calculate the equilbrium [MnO 4 ] (from outside so x not 2x ) in the solution.! (0.11 M) 7. Calculate the [OH ] (from outside ) needed to ppt Ni(OH) 2 so the [Ni +2 ]=1.0 x 10 8 M. What is the ph of this solution?! (2.3 x 10 4 M, ph = 10.37) Reaction quotient, Q, and Solubility (Q = K sp expression, not necessarily at equilibrium)! Q > K sp : Precipitation occurs until Q = K sp! Q = K sp : at Equilibrium, no net change, saturated solution exists! Q < K sp : Solid dissolves until Q = K sp 8. Will PbCl 2 ppt from a solution with a [Pb +2 ] = 1.0 x 10 4 M and a [Cl ]= 1.5 M? (Q= 2.3 x 10 4, Q>Ksp, yes ppt) 9. Will Mn(OH) 2 precipitate from a M solution of MnCl 2 if the ph is 8.00?! (Q= 5 x10 14, Q<Ksp, no ppt)

3 AP Chem: Complex Ion formation and the Formation constant, K f! Page 3 Ag + (aq) + 2 NH 3 (aq) < > Ag(NH 3 ) 2+ (aq)! K f = [Ag(NH 3 ) 2+ ] [Ag + ] [NH 3 ] 2 Notice now the ion concentrations are on the bottom, and everything is included!! K f values are large! So, usually almost all the metal ion you started with becomes the complex ion, so the equilibrium complex ion concentration is equal to the initial concentration of the metal ion. 1. You add concentrated NH 3 to 0.15 M AgNO 3 so the final [NH 3 ] is 1.5 M, and the final [Ag(NH 3 ) 2+ ] is 0.15 M. Calculate the final (or free) [Ag + ] in the solution.! (3.9x10 9 M)! b. If you add NaCl to this solution so the [Cl ] is M, would AgCl precipitate? (Hint: Calculate Q. Is it more or less than K sp for AgCl?)! (Q= 3.9x10 11, no ) 2. You add NH 3 to 1.0 x 10 3 M Cu(NO 3 ) 2 solution so the final [NH 3 ] is 0.10 M. Cu(NH 3 ) 4 + complex ions are made. Look up K f a. In the equilibrium solution, which is more concentrated, the free Cu +2 ions or the Cu(NH 3 ) 4 + complex ions? Why? b.!calculate the final [Cu +2 ] in the solution. (ignore any volume change)! (2x10 12 M) 3. Calculate the final [Cr +3 ] in a M Cr(NO 3 ) 2 solution with a buffered ph of (Hint: Cr(OH) 4 is made)! (1.9x10 16 M) 4. What ph do you need to reduce the [Zn +2 ] in a 0.10 M Zn(NO 3 ) 2 solution to M? (Hint: Zn(OH) 4 2 is formed)! (10.83)

4 Solubililty changes! Page 4 1. Which of the following salts would be more soluble in an acidic solution than pure water? (Hydroxides and Salts of weak acids dissolve more in acidic solutions. Salts of strong acids are unaffected.) a. Ag 2 CO 3! c. Ag 2 SO 4! e. PbCl 2 b. CeF 3! d. Cd(OH) 2! f. AgC 2 H 3 O 2 2. What happens in the following scenarios? Does the solid dissolve (D), a solid (or more solid) precipitate (ppt) form, or is there no change(nc)? Check your handout for complex ions and slightly soluble salts. If you add: a. NaI solution to AgNO 3 solution! b. NaCl solution to K 2 SO 4 solution c. Concentrated Na 2 SO 4 solution to a saturated solution of Ag 2 SO 4 d. Concentrated NaCl solution to a saturated solution of CaCO 3 e. Acid (H + ion) to Ca(OH) 2 solid in equilibrium with a saturated solution f. Base (OH ion) to Ca(OH) 2 solid in equilibrium with a saturated solution g. Concentrated NH 3 to solid CuCrO 4 in equilibrium with a saturated solution h. NH 3 to Fe 2 (SO 4 ) 3 solid in equilibrium with a saturated solution i. Acid to solid PbS 2 in equilibrium with a saturated solution j. Acid to solid PbCl 2 in equilibrium with a saturated solution k. A little base (OH ) to solid PbCl 2 equilibrium with a saturated solution! l. A lot of base (OH ) to solid PbCl 2 equilibrium with a saturated solution 3. In which solution is CaF 2 the most soluble? In which is it the least soluble?! a. plain water! b M HCl! c M KF! d M CaCl 2! 4. In which solution is PbSO 4 the most soluble? Least soluble?! a. plain water! b M HNO 3! c M NaNO 3! d. 1.0 M NH 3 5. In which solution is NiCO 3 the most soluble? Least soluble?! a. plain water! b M Na 2 CO 3! c. 1.0 M NH 3 (1. A, b, d, f more soluble in acid; 2a. AgI ppt, has K sp ; b. NC, all soluble; c. more ppt (add SO 4-2 common ion); d. NC (no com ion); e. D {OH dissove in acid}; f. more ppt {not an amph hydrox}; g. D {Cu(NH 3 ) 2 + is complex ion}; h. NC, no Fe(NH 3) complex ion; i. D, (salt of WA); j. NC (salt of SA); k. more ppt = Pb(OH) 2, l. D (Pb +2 = amp hydrox, add excess OH ); 3. b most, c least; 4. all same; 5. c most, b least)

5 AP Chem: Review for Ksp, Kf: Write the chemical equation and K expression first! Page 5 1a. Calculate the molar solubility of Ag 2 SO 4 in plain water.! (0.015 M)! b. Calculate the molar solubility of Ag 2 SO 4 in 0.10 M AgNO 3.! (1.4x10 3 M)! c. Why does the Molar solubility change between part a and b? 2. Calculate the molar solubility of Mn(OH) 2 in a buffer solution with ph = 9.00.! (1.9x10 3 M) 3a. The molar solubility of Sn(OH) 2 is 3.3 x M. Calculate K sp.! (1.4x10 28 )! b. NaOH is added to some Sn(NO 3 ) 2 solution. Sn(OH) 2 precipitates. The equilibrium [Sn +2 ] is 1.4 x M. Calculate the equilibrium [OH ].! (1.0 x 10 6 M)! c. What s the ph of the solution in part b?! (ph =8.0) 4a. You have 1.00 L of an HCN/NaCN buffer solution with a ph of 9.50 and a [HCN] of 0.20 M. Calculate the [CN ] in the buffer. (continued on back)! ([CN ]=0.31 M)

6 b. You add 1.0 x 10 4 mol of Ni +2 to this solution. Calculate the final [Ni +2 ]. Ni(CN) 2 4 is made. ([Ni +2 ] =5.4 x10 34 M) c. NaIO 3 is added so the [IO 3 ] is 0.25 M. Does Ni(IO 3 ) 2 ppt? Explain, show work. (Q= 3.4 x 10 35, so Q<Ksp, no ppt) p1. 3b. [OH ] is 2 times the Molar solubility, or 4.0 x 10 6 M, poh = 5.4, ph = 8.60! 4b. Common ion. Higher outside [I ] makes more ppt! 5a. K sp = x (2x) 2 = 4x 3 : [OH ] is from solid dissolving ( plain water ), so 2x! 5b. same as 3b, [OH ] is 2 times Molar solubility, or 3.8 x 10 5 M, poh = 4.42, ph = 9.58 p2. 5c. K sp = x (10 5 ) 2 :the poh=5, so [OH ]=10 5 M, (not 2x10 5 ) [OH ] is given so no times 2! 5d. same as 5c, [OH ] = 10 7 M, e. different [OH ] affect how much dissolves, but not K sp! 6. K sp = (2.0 x 10 8 ) x 2 : [MnO 4 ] is from outside so no times 2 :, not (2.0 x 10 8 )(2x) 2! 7. K sp = 1.7 x 10 8 (x) 2 and x = [OH ], from outside, not from solid dissolving, no 2x! 8. Q= 1x10 4 (1.5) 2 = 2.3 x10 4, K sp = 1.6 x10 5, [Cl ] given, so no times 2! 9. Q = (1 x 10 6 ) 2 [OH ] is given, no times 2 p 3 1. K f = 0.15! 1b. Q = (0.01) ( 3.9x10 9 ) x (1.5) 2!! 2a. K f is very big, so mostly products, Cu(NH 3 ) + 4! 2b. K f = 1.0 x 10 3! 3. K f = 0.015! 4. K f = 0.1 x (0.10) 4! x (10 4 ) 4! x 4 p a. K sp = 4x 3 plain water, Ag + from solid dissolving, so yes 2x! b. K sp = (0.10) 2 x Ag + is given, no times 2, [SO 2 4 ] = x = also solubility of Ag 2 SO 4! 2. K sp = x (10 5 ) 2 [OH - ] is given, no times 2! 3a. K sp = (3.3 x )(6.6 x ) 2 [OH ] from solid dissolving, so yes, [OH ]=2 x solubility! 3b.K sp = 1.4x10 16 (x 2 ) : [OH ] from outside so no 2x ; 3c. same idea, [OH ] = x! 4. Buffer equation : 3.2 x = Ka (0.20) [CN ] = 0.31 [CN ]! p. 6b. K f = 1.0 x 10 4 x (0.31) 4