Ksp_Mastery_Assignment_KEY

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Name:__________________________ Date:________________Period:_____ K sp Mastery Assignment **NOT necessarily AP style questions** 1. For slightly soluble salts, if Q < K is the solution saturated, unsaturated, or supersaturated? Justify your answer. If Q < K then the solution is unsaturated. Solution can accommodate more solute ions until equilibrium is reached. 2. For slightly soluble salts, if Q > K, would the mass of the precipitate increase or decrease? Justify your answer. If Q > K then the solution is supersaturated. Solution is unable to accommodate more solute ions and while salt will “crash out” as a precipitate. 3. Write the dissociation reactions and K sp expressions for the following slightly soluble salts. Write K sp in terms of the solubility, S. The first one has been done for you as a model. SALT DISSOCIATION REACTION EQUILIBRIUM EXPRESSION K sp in terms of “S” Cd(OH) 2 Cd(OH) 2 (s) Cd 2+ (aq) + 2OH (aq) K sp = [Cd 2+ ][ OH ] 2 K sp = 4S 3 CoCO 3 CoCO 3 (s) Co 2+ (aq) + CO 3 2- (aq) K sp = [Co 2+ ][CO 3 2- ] K sp = S 2 LaF 3 LaF 3 (s) La 3+ (aq) + 3 F - (aq) K sp = [La 3+ ]F - ] 3 K sp = 27S 4 Hg 2 S Hg 2 S (s) 2 Hg + (aq) + S 2- (aq) K sp = [Hg + ] 2 [S 2- ] K sp = 4S 3 Ba 3 (PO 4 ) 2 Ba 3 (PO 4 ) 2 (s) 3 Ba 2+ (aq) + 2 PO 4 3- (aq) K sp = [Ba 2+ ] 3 [PO 4 3- ] 2 K sp = 108S 5 4. Complete each calculation: a. Calculate the molar solubility for Cd(OH) 2 (K sp = 2.5x10 -14 ) and Ba 3 (PO 4 ) 2 (K sp = 3.4x10 -23 ). Cd(OH) 2 (s) Cd 2+ (aq) + 2OH (aq) K sp = [Cd 2+ ][OH - ] 2 2.5x10 -14 = [S][2S] 2 2.5x10 -14 = 4S 3 S = 1.8x10 -5 M Ba 3 (PO 4 ) 2 (s) 3 Ba 2+ (aq) + 2 PO 4 3- (aq) K sp = [Ba 2+ ] 3 [PO 4 3- ] 2 3.4x10 -23 = [3S] 3 [2S] 2 3.4x10 -23 = =108S 5 S = 1.56x10 -5 M b. The concentration of Pb 2+ in a solution saturated with PbBr 2 (s) is 2.2x10 -4 M, Calculate K sp for PbBr 2 . PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq) K sp = [Pb 2+ ][Br - ] 2 K sp = [2.2x10 -4 ][2(2.2x10 -4 )] 2 = 4.26x10 -11 AP Chemistry K sp Mastery Assignment 1
c. Calculate the solubility of solid Ca 3 (PO 4 ) 2 (K sp = 1.3x10 -32 ) in a 0.20 M Na 3 PO 4 solution. Ca 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 Ca 3 (PO 4 ) 2 Ca 2+ PO 4 3- Change 0 3S 2S + 0.20 1.3x10 -32 = [3S] 3 [2S + 0.20] 2 Assume S << 0.20 therefore 2S + 0.20 = 0.20 1.3x10 -32 = [3S] 3 [0.20] 2 S = 2.25x10 -11 d. The solubility of Ce(IO 3 ) 3 in a 0.20 M KIO 3 solution is 4.4x10 -8 mol/L. Calculate K sp for Ce(IO 3 ) 3 . Ce(IO 3 ) 3 (s) Ce 3+ (aq) + 3 IO 3 - (aq) K sp = [Ce 3+ ][IO 3 - ] 3 K sp = [Ce 3+ ][IO 3 - ] 3 Ce(IO 3 ) 3 Ce 3+ IO 3 - Change 0 S 3S + 0.20 K sp = [Ce 3+ ][IO 3 - ] 3 K sp = [4.4x10 -8 ][0.20 + 3(4.4x10 -8 )] 3 = 3.52x10 -10 e. Will a precipitate form when 75.0 mL of 0.020 M BaCl 2 and 125 mL of 0.040 M Na 2 SO 4 are mixed together? Use a calculation to justify your response. BaCl 2 (s) Ba 2+ (aq) + 2 Cl - (aq) [Ba 2+ ] = 0.075 L x 0.020 mol / 1 L = 0.0015 mol / 0.200 L = 0.0075 M Na 2 SO 4 (s) 2 Na + (aq) + SO 4 2- (aq) [SO 4 2- ] = 0.125 L x 0.040 mol / 1 L = 0.005 mol / 0.200 L = 0.025 M BaSO 4 (s) SO 4 2- (aq) + Ba 2+ (aq) K sp BaSO 4 = [0.0075][0.025] = 1.875x10 -4 K sp BaSO 4 = 1.08x10 -10 (LOOK UP VALUE) Q >> K therefore barium sulfate will precipitate f. A solution contains 1.0x10 -5 M Na 3 PO 4 . What is the minimum concentration of AgNO 3 that would cause precipitation of solid Ag 3 PO 4 ? (K sp = 1.8x10 -18 ) Na 3 PO 4 (s) 3 Na + (aq) + PO 4 -3 (aq) [PO 4 3- ] = 1.0x10 -5 M AgNO 3 (s) Ag + (aq) + NO 3 - (aq) [Ag + ] = X M Ag 3 PO 4 (s) PO 4 3- (aq) + 3Ag + (aq) K sp Ag 3 PO 4 = 1.8x10 -18 = [Ag + ] 3 [1x10 -5 ] [Ag + ] = 5.65x10 -5 Precipitate will form when [Ag + ] > 5.65x10 -5 g. In the above example: Will the addition of Na 3 PO 4 increase or decrease the solubility of barium phosphate at a given temperature? Justify your answer. AP Chemistry K sp Mastery Assignment 2
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