WA 08_CHE-122-jan18 sterile

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Feb 20, 2024

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Written Assignment 8: Solubility Equilibria Answer all assigned questions and problems, and show all work. 1. Silver chloride ( K sp = 1.6 × 10 –10 ) has a larger K sp than silver bromide ( K sp = 7.7 × 10 – 13 ). Which of these silver compounds will be the most soluble? (5 points) a. Silver chloride (larger K sp ) 2. Calculate the concentration of ions in the following saturated solutions: (a) [I – ] in AgI solution with [Ag + ] = 9.1 × 10 –9 M , (b) [Al +3 ] in Al(OH) 3 solution with [OH – ] = 2.9 × 10 –9 M . (20 points) a. K sp =(8.3x10 -17 ) = (9.1x10 -9 )[I - ], I - = 9.12x10 -9 b. K sp =(1.8x10 -33 ) = (2.9x10 -9 ) 3 [Al +3 ], Al +3 = 7.38x10 -8 (Reference: Chang 16.53) 3. The molar solubility of MnCO 3 is 4.2 × 10 –6 M . What is the K sp for this compound? (15 points) MnCO 3 ↔ Mn + + CO 3 - -s +s +s K sp = [Mn + ][CO 3 - ] = s 2 = (4.2x10 -6 ) 2 = 1.764x10 -11 (Reference: Chang 16.55) 4. What is the pH of a saturated zinc hydroxide solution? (15 points) K a =1.8x10 -14 = 4s 3 s = 1.65x10 -5 [OH] = 2s = 3.3x10 -5 pOH = -log (3.3x10 -5 ) = 4.48, pH = 9.5 (Reference: Chang 16.59) 5. If 20.0 mL of 0.10 M Ba(NO 3 ) 2 are added to 50.0 mL of 0.10 M Na 2 CO 3 , will BaCO 3 precipitate? (15 points) (0.1M)(0.02L)=(2x10 -3 mol)/0.07L = 2.85x10 -2 Ba (0.1M)(0.05L)=5x10 -3 mol/007L = 7.14x10 -2 CO 3 (2.85x10 -2 M Ba + )(7.14x10 -2 M CO 3 ) = 2.03x10 -3.p = Q K sp = 8.1 10 -9 Q>K sp , solution is supersaturated, BaCO 3 will precipitate (Reference: Chang 16.61) 6. Calculate the molar solubility of AgCl in a 1.00-L solution containing 10.0 g of dissolved CaCl 2 . (15 points) (10g)/(110.98g/mol) = 9.01x10 -2 /1L = 9.01x10 -2 M CaCl 2 [Cl - ] = 2(9.01x10 -2 ) = 0.1802 M [Ag + ][Cl - ] = K sp [Ag + ] = x, Total [Cl - ] = x+0.1802, 1 Copyright © 2017 by Thomas Edison State University. All rights reserved.
1.6x10 -10 = (0.1802)(x) X = 8.88x10 -10 = [Ag + ] (Reference: Chang 16.69) 7. Compare the molar solubility of Mg(OH) 2 in water and in a solution buffered at a pH of 9.0. (15 points) Mg(OH) 2 ↔ Mg 2+ + 2OH - M -s +s +2s (M-s) +s +2s K sp = s(2s) 2 = 4s 3 = 1.2x10 -11 , s = 1.44x10 -4 in water 14-9 = 5 = pOH, 10 -5 = 1.0x10 -5 (OH buffered) Mg(OH) 2 ↔ Mg 2+ + 2OH - M 1.0x10 -5 (buffered) -s +s +2s (M-s) +s 1.0x10 -5 +2s (buffered) K sp = s(1.0x10 -5 ) 2 = 1.2x10 -11 S = 0.12M **solubility is much higher (Reference: Chang 16.73) 2 Copyright © 2017 by Thomas Edison State University. All rights reserved.
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