Chem Solubility Quiz Practice

CHEM SOLUBILITY QUIZ Practice Using K trial (or “Q”) to determine if a solution is supersaturated, saturated, or unsaturated. 1. A solution is created by adding 0.35 moles of zinc sulphide solid to 1.0 L of water, and heating it gently on a hot plate to promote dissolving. A) Is this a supersaturated, saturated, or unsaturated solution? B) State the concentrations of each ion at equilibrium. C) If a PPT forms, show the calculations to determine the mass of PPT that forms. 2. Lanthanum iodate, La(IO3)3 has a Ksp = 7.5 x 10-12. If a chemist dissolves 1.8 X 10-4 moles of solute in 500.0 ml of water: A) Is this a supersaturated, saturated, or unsaturated solution? B) Which way will this reaction proceed to reach equilibrium (or saturation)? C) If a PPT forms, show the calculations to determine the mass of PPT that forms. 3. A solution of Magnesium Hydroxide is found to have a [OH- ] of 2.24 x 10-4 M. A) Is this a supersaturated, saturated, or unsaturated solution? B) State the concentrations of each ion at equilibrium. C) If a PPT forms, show the calculations to determine the mass of PPT that forms. 4. A solution of BaCO3 is created such that it’s concentration is determined to be 2.6 x 10-3 M. A) Is this a supersaturated, saturated, or unsaturated solution? B) State the concentrations of each ion at equilibrium. C) If a PPT forms, show the calculations to determine the mass of PPT that forms. ANSWERS 1. A) supersaturated because Ksp < Ktrial B) the concentrations of each ion at equilibrium: approx. 0 M C) the mass of PPT that forms: 34 g 2. A) unsaturated solution because Ksp > Ktrial B) To reach equilibrium (or saturation): SHIFTS TO THE RIGHT (+x for the products); i.e. more solute could be added to reach the maximum amount that can dissolve in solution 4 I didn’t ask for the final concentrations because Ksp = 27s . -12 -4 -4 3 This is pretty tedious algebra to solve: 7.5 x 10 = (3.6 X 10 +x) • (3(3.6 X 10 )+3x) C) NO PPT FORMS 3. A) saturated solution because Ksp = Ktrial -4 -4 +2 B) concentrations of each ion at equilibrium: [OH- ] = 2.24 x 10 M = 2s So, s = 1.12 x 10 M = [Mg ] C) NO PPT forms (however if I am looking at a beaker full of saturated solution, I might see some white specks of solid floating around in my aqueous solution). 4. A) supersaturated because Ksp < Ktrial -3 -4 B) concentrations of each ion at equilibrium 2.6 x 10 M – x = 1.0 x 10 M C) mass of PPT that forms = 0.49 g CHEM SOLUBILITY QUIZ Practice FULL ANSWER KEY: Using K trial (or “Q”) to determine if a solution is supersaturated, saturated, or unsaturated. 1. ⇌ ZnS (s) I C E Zn+2 (aq) 0.35 moles of solute 0.35 M mixed into 1.0 L of solution Ksp < K trial 2.0 x 10-25 < (0.35)2 SUPERSATURATED! ß shifts to the left -x “x” moles of Precipitate forms 0.35 – x + S-2 (aq) 0.35 M -x 0.35 - x Ksp = 2.0 x 10-25 = (0.35 – x)2 √ both sides 4.5 x 10-13 = 0 .35 – x x = 0.35 M [Zn+2] = [S-2] = 0 M 0.35 moles / L × 1.0 L × 97.5 g / mole = 34 g of PPT forms 2. I C ⇌ La(IO3)3 (s) -4 1.8 x 10 moles of solute mixed into 500.0 mL of solution = 3.6 x 10-4 La+3 (aq) = 3.6 x 10-4 M + 3IO3- (aq) =3(3.6 x 10-4) =0.00108 M K trial = 27s4 = 27(3.6 x 10-4)4 = 4.5 x 10 -13 Ksp > K trial 7.5 x 10-12 > 4.5 x 10-13 UNSATURATED! shifts to the right to achieve equilibrium à i.e. “x” more solute could be added to reach the maximum amount that can dissolve in solution +x +3 x E No PPT forms. I did not ask you to solve for the final concentration of ions. To do that, you 3 would have to solve 7.5 x 10-12 =(3.6 x 10- 4 +x)[ 3(3.6 x 10-4) +3x] which is a little too tedious. CHEM SOLUBILITY QUIZ Practice FULL ANSWER KEY: Using K trial (or “Q”) to determine if a solution is supersaturated, saturated, or unsaturated. 3. ⇌ Mg(OH)2 (s) Mg+2 (aq) + 2OH - (aq) 2.24 x 10-4 2s s 1.12 x 10-4 K trial = 4s3 =4(1.12 x 10-4)3 = 5.62 x 10 -12 Ksp = K trial 5.6 x 10 -12 = 5.62 x 10 -12 SATURATED SOLUTION! Already at equilibrium !!! If the solution is changed AT ALL, (for example, if you allow water to evaporate or allow the temperature of the solution to change, the equilibrium will shift left (PPT will occur) or right (there will no longer be maximum solute dissolved in solution) Ion concentrations as shown above. 4. ⇌ BaCO3 (s) I C E Ba+2 (aq) 2.6 x 10-3 M + CO3-2 (aq) 2.6 x 10-3 M K trial = (2.6 x 10-3)2 = 6.76 x 10-6 Ksp < K trial 2.6 x 10-9 < 6.76 x 10-6 SUPERSATURATED! ß shifts to the left -x -x “x” moles of Precipitate forms 2.6 x 10-3 – x 2.6 x 10-3 - x Ksp = 2.6 x 10-9= (2.6 x 10-3 – x)2 √ both sides 5.1x 10-5 = 2.6 x 10-3 – x x = 2.5 x 10-3 M [Ba+2] = [CO3-2] = 2.6 x 10-3 – x = 2.6 x 10-3 – 2.5 x 10-3 = 0.0001 M x = 2.5 x 10-3 2.5 x 10-3 moles ×197.3 g / mole = 0.49 g of PPT forms in the 1.0 L solution beaker. The precipitate can be filtered out and the remaining solution will be a saturated solution.