Solubility Workbook Questions

Power Point Lesson Notes- double click on the lesson number

Worksheets Quiz

1. Solubility and Saturated Solutions. WS 1 1

2. Ion Concentration Calculations and Ionic Equations. WS 2 2

3. Solubility to Ksp. WS 3 3

4. Ksp to Solubilty and Size of Ksp. WS 4 4

5. Trial Ksp. WS 5 5

6. Separating Ions WS 6 6

7. Common Ion Effect and WS 7 7

8. Titrations and Max Ion Concentration   WS 8 Web Review 8 Quizmebc

9. Review Practice Test1

10. Review Practice Test 2

11. Test

Text book Hebden Read Unit III

The following workbook is designed to ensure that you can demonstrate your understanding of all aspects of the solubility unit. Ask yourself, “do I want to do well in this class?” If you are determined to be successful the minimum expectation that you should have for yourself is that you do all of these questions by the due dates given by your teacher. There are other things that you should do to prepare for the test at the end of the unit. Remember, what you put into this course is what you will get out. There is no substitute for consistent effort and hard work. If you can’t do a question, get some help before the end of the unit, you need to know, understand, and remember everything. Good luck! I know you can do well in this unit. Keep up the great work!

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Chemistry 12 Solubility and Saturated Solutions WS #1

1. Define and give units for solubility. M, g/100mL, g/L

2. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to an unsaturated solution.

Rate of dissolving is greater than the rate of crystallization

3. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to a saturated solution.

Rate of dissolving equals the rate of crystallization

4. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to a supersaturated solution.

Rate of dissolving is less than the rate of crystallization

5. Which of the above solutions would need to be prepared in order to determine the solubility of an ionic solution.

 Saturated

6. 2.65 g of Ba(OH)₂ is dissolved in 70.0 mL of water to produce a saturated solution at 20 ^oC. Calculate the solubility in units of g/100 mL, g/L, and M.

0.221M

37.9g/L

3.79g/100mL

7. A beaker containing 100. mL of saturated BaCO₃ solution weighs 159.60 g. The beaker is evaporated to dryness and weighs 56.36 g. The empty beaker weighs 24.33 g. Calculate the solubility in units of g/100 mL, g/ L, and M.

56.36 g 32.03g = 320g x 1 mole = 1.62 M Do not use the mass with water included!

- 24.33g 100 mL L 197.3g

32.03 g

8. Write dissociation equations to represent the equilibrium present for a saturated solution of each ionic compound. Write the solubility product (Ksp expression) for each of the equilibrium systems. The first one is done.

a) Al₂(SO₄)₃ ⇄ 2Al³⁺ + 3SO₄^2-   Ksp = [Al³⁺]² [SO₄^2-]³

b) FeCO₃ ⇄ Fe²⁺ + CO₃^2-  Ksp = [Fe²⁺] [CO₃^2-]

c) Co₂(SO₄)₃ ⇄ 2Co³⁺ + 3SO₄^2-    Ksp = [Co³⁺]² [SO₄^2-]³

d) Na₃PO₄ ⇄ 3Na⁺ + PO₄^3-  Ksp = [Na⁺]³ [PO₄^3-]

10. Write formula, complete ionic, and net ionic equations for each.

a) H₂SO_4(aq) + NaOH_(aq) →

H₂SO_{4 (aq)} + 2NaOH _(aq) → 2H₂O _(l) + Na₂SO_{4 (aq)}

2H⁺_(aq) + SO₄^2-_(aq) + 2Na⁺_(aq) + 2OH^-_(aq) → 2H₂O _(l) + 2Na⁺_(aq) + SO₄^2-_(aq)

2H⁺_(aq) + 2OH^-_(aq) → 2H₂O _(l)

b) Mg(NO₃)_2(aq) + Na₂CO_3(aq) →

 Mg(NO₃)_{2 (aq)} + Na₂CO_{3 (aq)} → MgCO_{3 (s)} + 2NaNO_{3 (aq)}

Mg²⁺_(aq) + 2NO₃^-_(aq) + 2Na⁺_(aq) + CO₃^2-_(aq) → MgCO_{3 (s)} + 2Na⁺_(aq) + 2NO₃ ^- _(aq)

Mg²⁺_(aq) + CO₃^2-_(aq) → MgCO_{3 (s)}

 c) Al(NO₃)_3(aq) + (NH₄)₃PO_4(aq) →

Al(NO₃)_{3 (aq)} + (NH₄)₃PO_{4 (aq)} → AlPO_{4 (s)} + 3NH₄NO_{3 (aq)}

Al³⁺_(aq) + 3NO₃^-_(aq) + 3NH₄⁺_(aq) + PO₄^3-_(aq) → AlPO_{4 (s)}+ 3NH₄⁺_(aq) + 3NO₃ ^- _(aq)

Al³⁺_(aq) + PO₄^3-_(aq) → AlPO_{4 (s)}

 d) H₃PO_4(aq) + Ca(OH)_2(aq) →

2H₃PO_4(aq) + 3Ca(OH)_2(aq) → Ca₃(PO₄)_2(s) + 6HOH_(l)

6H⁺ _(aq) + 2PO₄^3-_(aq) + 3Ca²⁺_(aq) + 6OH^-_(aq) → Ca₃(PO₄)_2(s) + 6HOH_(l)

6H⁺ _(aq) + 2PO₄^3-_(aq) + 3Ca²⁺_(aq) + 6OH^-_(aq) → Ca₃(PO₄)_2(s) + 6HOH_(l)

Chemistry 12 Solubility WS #2 Ion Concentration Calculations

1. What is the concentration of each ion in a 10.5 M sodium silicate solution?

 Na₂SiO₃ ⇄ 2Na⁺ + SiO₃^2-

10.5 M 21.0 M 10.5 M

[Na⁺] = 21.0 M, [SiO₃^2-] = 10.5 M

2. What is the concentration of each ion in the solution formed when 94.5 g of nickel (III) sulphate is dissolved into 850.0 mL of water?

 Molarity = 94.5 g x 1 mole

405.7g = 0.2740

0.850L

Ni₂(SO₄)₃ ⇄ 2Ni³⁺ + 3SO₄^2-

0.2740 0.548 M 0.822 M

[Ni³⁺] = 0.548 M, [SO₄^2-] = 0.822 M

3. If 3.78 L of 0.960 M sodium fluoride solution is added to 6.36 L of 0.550 M calcium nitrate solution, what is the resulting concentration of [Ca⁺²] and [F^-]?

NaF ⇄ Na⁺ + F^- Ca(NO₃)₂ ⇄ Ca²⁺ + 2NO₃^-

3.78 x 0.960 M = 0.358 M 0.358M 6.36 x 0.550 M = 0.345 M 0.690M

10.14 10.14

[Ca²⁺] = 0.345 M, [F^-] = 0.358 M

4. What is the concentration of each ion in the solution formed when 94.78 g of iron (III) sulphate is dissolved into 550.0 mL of water?

[Fe³⁺] = 0.8619 M, [SO₄^2-] = 1.293 M

5. If the [F^-] = 0.200 M, calculate the number of grams AlF₃ that would be dissolved in 2.00 L of water.

AlF₃ ⇄ Al³⁺ + 3F^-

0.06667M 0.06667M 0.200M

2.00L x 0.06667 mole x 84.0 g = 11.2g

L mole

6. If the [SO₄^2-] = 0.200 M in 2.0 L of Al₂(SO₄)₃, determine the [Al³⁺] and the molarity of the solution.

Al₂(SO₄)₃ ⇄ 2Al³⁺ + 3SO₄^2-

0.067 M 0.13 M 0.20 M

Dissociation Equations Write a dissociation equation for any chemical which dissociate when dissolved in water:

1. HCl _(aq) ⇄ H⁺ + Cl^-

            2. C₆H₁₂O_{6 (s)} ⇄ C₆H₁₂O_6(aq) (molecular compounds do not dissociate)

3. Na₂S _(s) ⇄ 2 Na⁺ _(aq) + S^2- _(aq)

4. Al(CH₃COO)_{3 (s)} ⇄ Al³⁺ _(aq) + 3CH₃COO^- _(aq)

5. MgBr_{2 (s)} ⇄ Mg²⁺ _(aq) + 2 Br^- _(aq)

6. Na₂CO_{3 (s)} ⇄ 2 Na⁺ _(aq) + CO₃^2- _(aq)

7. C₁₂H₂₂O_{11 (s)} ⇄ C₁₂H₂₂O_{11 (aq)} (molecular compounds do not dissociate)

8. K₃PO_{4 (s)} ⇄ 3 K⁺ _(aq) + PO₄^3- _(aq)

9. CH₃OH _(l) ⇄ CH₃OH _(aq) (molecular compounds do not dissociate)

Net Ionic Equations

Write chemical equations, total ionic equations and net ionic equations for each reaction. The first one is done for you. (assume that all reactions occur):

1. Magnesium metal is placed in hydrochloric acid

Mg _(s) + 2 HCl _(aq) → MgCl_{2 (aq)} + H_{2 (g)}

Mg _(s) + 2 H⁺ _(aq) + 2 Cl^- _(aq) → Mg²⁺ _(aq) + 2Cl^- _(aq) + H_{2 (g)}

Mg _(s) + 2 H⁺ _(aq) → Mg²⁺ _(aq) + H_{2 (g)}

2. Zinc metal is placed in silver nitrate solution

Zn _(s) + 2Ag NO_{3 (aq)} → Ag _(s) + 2 Zn(NO₃)_{2 (aq)}

Zn_(s) + 2Ag⁺_(aq) + 2 NO₃^-_(aq) → 2Ag_(s) + Zn⁺² _(aq) + + 2NO₃^-_(aq)

Zn_(s) + 2Ag⁺ _(aq) → Zn⁺² _(aq) + 2Ag_(s)

3. Barium chloride solution is added to lead (II) nitrate solution.

BaCl_{2 (aq)} + Pb(NO₃)_{2 (aq)} → PbCl_{2 (s)} + Ba(NO₃)_{2 (aq)}

Ba²⁺_(aq) + 2Cl^-_(aq) + Pb²⁺_(aq) + 2NO₃^-_(aq) → PbCl_{2 (s)} + Ba²⁺ _(aq) + 2NO₃ ^- _(aq)

Pb²⁺ _(aq) + 2Cl^- _(aq) → PbCl_{2 (s)}

4. Sulphuric acid is added to Strontium hydroxide solution.

H₂SO_{4 (aq)} + Sr(OH)_{2 (aq)} → 2H₂O _(l) + SrSO_{4 (s)}

2H⁺_(aq) + SO₄^2-_(aq) + Sr²⁺_(aq) + 2OH^-_(aq) → 2H₂O _(l) + SrSO_{4 (s)}

2H⁺_(aq) + SO₄^2-_(aq) + Sr²⁺_(aq) + 2OH^-_(aq) → 2H₂O _(l) + SrSO_{4 (s)}

5. Sodium carbonate solution is added to nickel (III) nitrate solution.

3Na₂CO_3(aq) + 2Ni(NO₃)_{3 (aq)} → Ni₂(CO₃)_{3 (s)} + 6NaNO_{3 (aq)}

6Na⁺_(aq) +3CO₃^2-_(aq) +2Ni³⁺_(aq) +6NO₃^-_(aq) → Ni₂(CO₃)_{3 (s)} + 6Na⁺_(aq) +6NO₃^-_(aq)

3CO₃^2-_(aq) + 2Ni³⁺_(aq) → Ni₂(CO₃)_{3 (s)}

 6. Aqueous chlorine is added to sodium bromide solution.

Cl_{2 (aq)} + 2NaBr _(aq) → 2NaCl _(aq) + Br_{2 (aq)}

Cl_{2 (aq)} + 2Na⁺ _(aq) + 2Br^- _(aq) → 2Na⁺ _(aq) + 2Cl^- _(aq) + Br_{2 (aq)}

Cl_{2 (aq)} + 2Br^- _(aq) → 2Cl^- _(aq) + Br_{2 (aq)}

 7. Nitric acid is added to aluminum hydroxide solution.

2HNO_{3 (aq)} + Sr(OH)2 _(aq) → 2H₂O _(l) + Sr(NO₃)_{2 (aq)}

2H⁺_(aq) + 2NO₃^-_(aq) + Sr²⁺_(aq) + 2OH^-_(aq) → 3H₂O _(l) + Sr²⁺_(aq) + 2NO₃^-_(aq)

H⁺_(aq) + OH^-_(aq) → H₂O _(l)

Chem 12 WS #3 Solubility to K_sp

The K_sp is a measure of the solubility of an ionic salt. The larger the value of the K_sp, the greater is the solubility of the salt. You can only calculate a K_sp if the solution is saturated. Only saturated salt solutions are in equilibrium. You can calculate the K_sp from the solubility of a salt, since the solubility represents the concentration required to saturate a solution.

1. Calculate the K_sp for CaCl₂ if 200.0 g of CaCl₂ are required to saturate 100.0 mL of solution.

Molarity = 200 g x 1 mole

111.1 g = 18.001 M

0.1000 L

CaCl₂ ⇄ Ca²⁺ + 2Cl^-

s                                   s                         2s

Ksp = [Ca²⁺][Cl^-]²

Ksp = 4s³

Ksp = 4(18.001)³

Ksp = 2.333 x 10⁴

2. Calculate the K_sp for AlCl₃ if 100.0 g is required to saturate 150.0 mL of a solution.

Ksp = 1.679 x 10⁴

3. The solubility of SrF₂ is 2.83 x 10^-5 M. Calculate the K_sp.

Ksp = 9.07 x 10^-14

4. The solubility of GaBr₃ is 15.8 g per 100 mL. Calculate the K_sp.

Molarity = 15.8 g x 1 mole

309.4 g = 0.51066 M

0.100 L

GaBr₃ ⇄ Ga³⁺ + 3Br^-

s s                       3s

Ksp = [Ga³⁺][Br^-]³

Ksp = 27s⁴

Ksp = 27[0.51066]⁴

Ksp = 1.84

5. The solubility of Ag₂SO₄ is 1.33 x 10^-7g per 100 mL. Calculate the K_sp.

Ksp = 3.10 x 10^-25

6. If 2.9 x 10^-3 Ca(OH)₂ g is needed to saturate 250 mL of solution, what is the K_sp.

Ksp =1.5 x 10^-11

7. At a certain temperature, a 40.00 mL sample of a saturated solution of barium hydroxide, is neutralized by 29.10 mL of 0.300 M HCl. Calculate the Ksp of Ba(OH)₂.

2HCl + Ba(OH)₂ → BaCl₂ + 2HOH

0.02910 L 0.0400 L

0.300 M ? M

Molarity Ba(OH)₂ = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)₂

1 L 2 moles HCl

0.0400 L

= 0.109 M

Ba(OH)₂ ⇄ Ba²⁺ + 2OH^-

               s                              s                                     2s            

Ksp = [Ba²⁺][OH^-]²

Ksp     = 4s³

          Ksp      = 4(0.109³

= 5.20 x 10^-3

Calculate the concentrations of all ions in each solution.

8. 0.50 M Al₂(SO₄)₃(aq)

Al₂(SO₄)₃ ⇄ 2Al³⁺ + 3SO₄^2-

0.50 M 1.0 M 1.5 M

[Al³⁺] = 1.0M [SO₄^2-] = 1.5M

9. 25.7g (NH₄)₃PO₄ (aq) in 250mL H₂O.

[NH₄⁺] = 2.07M [PO₄^3-] = 0.690M

10. 210g CoCl₂ • 6H₂O in 800mL H₂O.

[Co²⁺] = 1.10M [Cl^-] = 2.20M

Chemistry 12 Ksp to Solubility WS # 4

Calculate the solubility in M and g/L for each. Use the K_sp values found in your chart.

1) BaCO₃

BaCO_3(s) ⇄ Ba²⁺ + CO₃^2-

x x x

ksp = [Ba²⁺][ CO₃^2-]

ksp = x²

2.6 x 10^-9 = x²

5.099 x 10^-5 M

5.099 x 10^-5 mole x 197.3 g =

           L    v 1 mole

1.0 x 10^{-2 g}/_L

2) Fe(OH)₂

2.1 x 10^{-4 g}/_L

3) PbCl₂

4.0 ^g/_L

4) How many grams of Mg(OH)₂ are required to completely saturate 1.5 L of solution?

Mg(OH)₂ ⇄ Mg²⁺ + 2OH^-

s s 2s

Ksp = [Mg²⁺][OH^-]²

Ksp = [s][2s]²

Ksp = 4s³

5.6 x 10^-12 = 4s³

1.119 x 10^-4 M = s

1.5 L x 1.119 x 10^-4 mole x 58.3 g = 9.8 x 10^-3 g

1 L 1 mole

Review

1. If 200 g of MgCl₂ is required to saturate 1.5 L of solution at 20 ^oC, calculate the K_sp.

Ksp = 11

2. If the K_sp for Al₂O₃ is 2.8 x 10-8, calculate [Al³⁺] and [O^-2] in ^mol/_L.

[Al^+3] = 2.4 x 10^-2 M [O^-2] = 3.6 x 10^-2 M

Trial Ksp Worksheet 5

1. Will a precipitate form if 200ml 0.00020M Ca(NO₃)₂ is mixed 300ml of 0.00030M Na₂C0₃?

CaCO₃ ⇄ Ca²⁺ + CO₃^2-

200 x 0.00020 M 300 x 0.00030 M

500 500

0.000080 M 0.00018 M

Trial Ksp = [0.000080][0.00018]

Trial Ksp = 1.4 x 10^-8 > Ksp(5.0 x 10^-9)

Therefore a precipitate forms!

2. Will a precipitate form if 25.0ml of .0020M Pb(NO₃)₂ is mixed with 25.0ml of .040M NaBr.

Trial ksp = 4.0 x 10^-7 no ppt

3. Will a precipitate form if equal volumes of 0.00020M Ca(NO₃)₂ is mixed with 0.00030M Na₂C0₃?

 Note: When equal volumes are mixed the dilution factor is ½ for each ion.

Trial ksp = 1.5 x 10^-8 ppt > Ksp and there is a precipitate

Ksp

4. Co(OH)₂ Solubility = 3.0x10^-3 g/L Ksp=?

ksp = 1.3 x 10^-13

5. Ag₂C₂O₄ Solubility = 8.3x10^-4 M Ksp=?

ksp = 2.3 x 10^-9

Solubility

6. SrF₂ Solubility in (M) = ?

1.0 x 10^-3 M

7. Cu(IO₃)₂ Solubility (g/L) = ?

 1.1 g/L

Separation Positive Ions: Work from top to bottom of solubility chart!! WS # 7

1. Ag⁺ Mg²⁺ Ba²⁺

i) Add: NaCl_(aq) Filter Out: AgCl_(s) Net Ionic equation: Ag⁺ + Cl^- ------> AgCl_(s)

ii) Add: Na₂SO_4(aq) Filter Out: BaSO_4(s) Net Ionic equation: Ba²⁺ + SO₄^-2 ------> BaSO_4(s)

iii) Add: NaOH_(aq) Filter Out: Mg(OH)_2(s) Net Ionic equation: Mg²⁺+2OH^-------> Mg(OH)_2(s

2. Pb²⁺ Ba²⁺ Sr²⁺

i) Add: NaCl_(aq) Filter Out: PbCl_2(s) Net Ionic equation: Pb⁺² + 2Cl^- ------> PbCl_2(s)

ii) Add: NaOH_(aq) Filter Out: Ba(OH)_2(s) Net Ionic equation: Ba²⁺+2OH^-------> BaOH)_2(s)

iii) Add: Na₃PO_4(aq) Filter Out: Sr₃(PO₄)_2(s) Net Ionic equation: 3Sr²⁺+2PO₄^-3------> Sr₃(PO₄)_2(s)

3. Cu⁺ Ca²⁺ Sr²⁺

i) Add: NaCl_(aq) Filter Out: CuCl_(s) Net Ionic equation: Cu⁺ + Cl^- ------> CuCl_(s)

ii) Add: NaOH_(aq) Filter Out: Ca(OH)_2(s) Net Ionic equation: Ca²⁺+2OH^-------> Ca(OH)_2(s)

iii) Add: Na₃PO_4(aq) Filter Out: Sr₃(PO₄)_2(s) Net Ionic equation: 3Sr²⁺+2PO₄^-3------> Sr₃(PO₄)_2(s)

4. Be²⁺ Sr²⁺ Ag⁺

i) Add: NaCl_(aq) Filter Out: AgCl_(s) Net Ionic equation: Ag⁺ + Cl^- ------> AgCl_(s)

ii) Add: Na₂SO_4(aq) Filter Out: SrSO_4(s) Net Ionic equation: Sr²⁺ + SO₄^-2 ------> SrSO_4(s)

iii) Add: NaOH_(aq) Filter Out: Be(OH)_2(s) Net Ionic equation: Be²⁺+2OH^-------> Be(OH)_2(s)

5. Be²⁺ Ca²⁺ Pb²⁺

i) Add: NaCl_(aq) Filter Out: PbCl_2(s) Net Ionic equation: Pb⁺² + 2Cl^- ------> PbCl_2(s)

ii) Add: Na₂SO_4(aq) Filter Out: CaSO_4(s) Net Ionic equation: Ca²⁺ + SO₄^-2 ------> CaSO_4(s)

iii) Add: NaOH_(aq) Filter Out: Be(OH)_2(s) Net Ionic equation: Be²⁺+2OH^-------> Be(OH)_2(s)

]

6. Calculate the Ksp for CaCl₂, if 50.0 g is required to saturate 25.0 mL of water.

50.0 g x 1 mole

111.1 g = 18.0 M Ksp = 4s³ = 4(18.0)³ = 2.33 x 10⁴

0.0250 L

7. Calculate the molar solubility of Mg(OH)₂.

Mg(OH)₂ ⇄ Mg²⁺ + 2OH^-

s s 2s

4s³ = 5.6 x 10^-12

s = 1.1 x 10^-4 M

8. Will a precipitate form if equal volumes of 0.00020 M Na₂CO₃ is mixed with 0.00020 M MgCl₂.

MgCO_3(s) ⇄ Mg²⁺ + CO₃^2-

½ (0.00020 M) ½ (0.00020 M)

0.00010 M 0.00010 M

Trial Ksp = [Mg²⁺][CO₃^2-]

(0.00010)(.00010)

1 x 10^-8

Trial Ksp < Ksp = 6.8 x 10^-6 no ppt

9. Write the formula, complete, and net ionic equation.

Formula Equation: CaCl_2(aq) + 2AgNO_3(aq) → Ca(NO₃)_2(aq) + 2AgCl_(s)

Complete Ionic: Ca²⁺ + 2Cl^- + 2Ag⁺ + 2NO₃^- → Ca²⁺ + 2NO₃^- + 2AgCl_(s)

Net Ionic: Ag⁺ + Cl^- → AgCl_(s)

1. SO₃^2- OH^- I^-

i) Add: Sr(NO₃)_2(aq) Filter Out: SrSO_3(s) Net Ionic equation: Sr²⁺ + SO₃^2- ---->SrSO_3(s)

ii) Add: Zn(NO₃)_2(aq) Filter Out: Zn(OH)_2(s) Net Ionic equation: Zn²⁺ + 2OH^- ----> Zn(OH)_2(s)

iii) Add: AgNO_3(aq) Filter Out: AgI_(s) Net Ionic equation: Ag⁺ + I^- ---->AgI_(s)

2. CO₃^2- OH^-

i) Add: Sr(NO₃)_2(aq) Filter Out: SrCO_3(s) Net Ionic equation: Sr²⁺ + CO₃^2- ---->SrCO_3(s)

ii) Add: Zn(NO₃)_2(aq) Filter Out: Zn(OH)_2(s) Net Ionic equation: Zn²⁺ + 2OH^- ----> Zn(OH)_2(s)

3. Br^- S^2- PO₄^3-

i) Add: Sr(NO₃)_2(aq) Filter Out: Sr₃(PO₄)_2(s) Net Ionic equation: 3Sr²⁺ + 2PO₄^3- ----> Sr₃(PO₄)_2(s)

ii) Add: Zn(NO₃)_2(aq) Filter Out: ZnS_(s) Net Ionic equation: Zn²⁺ +  S^-2 ----> ZnS_(s)

iii) Add: AgNO_3(aq) Filter Out: AgBr_(s) Net Ionic equation: Ag⁺ + Br^- ---->AgBr_(s)

4. PO₄^3- OH^- S^2-

i) Add: Sr(NO₃)_2(aq) Filter Out: Sr₃(PO₄)_2(s) Net Ionic equation: 3Sr²⁺ + 2PO₄^3- ----> Sr₃(PO₄)_2(s)

ii) Add: Mg(NO₃)_2(aq) Filter Out: Mg(OH)_2(s) Net Ionic equation: Mg²⁺ + 2OH^- ----> Mg(OH)_2(s)

iii) Add: Zn(NO₃)_2(aq) Filter Out: ZnS_(s) Net Ionic equation: Zn²⁺ + S^-2 ----> ZnS_(s)

5. OH^- S^2- SO₄^2-

i) Add: Mg(NO₃)_2(aq) Filter Out: Mg(OH)_2(s) Net Ionic equation: Mg²⁺ + 2OH^- ----> Mg(OH)_2(s)

ii) Add: Ba(NO₃)_2(aq) Filter Out: BaSO_4(s) Net Ionic equation: Ba²⁺ + SO₄^2- ---->BaSO_4(s)

iii) Add: Zn(NO₃)_2(aq) Filter Out: ZnS_(s) Net Ionic equation: Zn²⁺ +  S^-2 ----> ZnS_(s)

6. S^2- SO₄^2- Cl^-

i) Add: Ba(NO₃)_2(aq) Filter Out: BaSO_4(s) Net Ionic equation: Ba²⁺ + SO₄^2- ---->BaSO_4(s)

ii) Add: Zn(NO₃)_2(aq) Filter Out: ZnS_(s) Net Ionic equation: Zn²⁺   +  S^-2 ----> ZnS_(s)

iii) Add: AgNO_3(aq) Filter Out: AgCl_(s) Net Ionic equation: Ag⁺ + Cl^- ---->AgCl_(s)

Common Ion Effect Worksheet # 8

Consider the following equilibrium system. PbCl_2(s) ⇌ Pb²⁺_(aq) + 2 Cl^-_(aq)

Describe what happens to the solubility of PbCl₂ after each of the changes are made.

Solubilty

1. PbCl_2(s) is added no change

2. Pb(NO₃)₂ is added decreases

3. NaCl is added decreases

4. H₂O is added (remember that solubility is moles per litre) no change

5. AgNO₃ is added (Ag⁺ reacts with Cl^- to form AgCl_(s) which has low solubility)                                         increases

6. NaBr is added (Br^- reacts with Pb²⁺ to form PbBr_2(s) which has low solubility)                                         increases

Consider the following equilibrium system. AgBr_(s) ------> Ag⁺_(aq) + Br^-_(aq)

Describe what happens to the solubility of PbCl₂ after each of the changes are made.

Solubilty

7. AgBr_(s) is added no change

8. Pb(NO₃)₂ is added Pb²⁺ reacts with Br^- to form PbBr_2(s) which has low solubility) increases

9. NaCl is added (Ag⁺ reacts with Cl^- to form AgCl_(s) which has low solubility)                                                                                                       increases

10. H₂O is added (remember that solubility is moles per litre)                                                    no change

11. AgNO₃ is added               decreases

12. NaBr is added decreases

13. Explain why more Zn(OH)_2(s) dissolves when 3 M HCl is added to a saturated solution of Zn(OH)₂.

            Start by writing the correct equilibrium equation.

Zn(OH)_2(S) ⇌ Zn²⁺ + 2OH^-

The HCl increases the concentration of H⁺ which reacts with OH^- lowering the [OH^-].

            This causes the above equilibrium to shift to the right and more Zn(OH)_2(S) dissolves.

14. In an experiment, 0.1 M AgNO₃ is added to 0.1 M NaCl, resulting in the formation of a white precipitate.

            When 0.1 M NaI is added to this mixture, the white precipitate dissolves and a yellow precipitate forms.

The formula for the white precipitate is AgCl

The formula for the yellow precipitate is AgI

The net ionic equation for the first equilibrium is AgCl_(s) ⇌ Ag⁺ + Cl^-

The net ionic equation for the formation of the yellow precipitate is              Ag⁺ + I^- → AgI_(s)

Explain why the white precipitate dissolves. Start by writing the equilibrium equation for the white precipitate, then,

            explain how adding NaI affects this equilibrium.

AgCl_(s) ⇌ Ag⁺ + Cl^-

The NaI increases the concentration of I^- which reacts with Ag⁺ lowering the [Ag⁺].

            This causes the above equilibrium to shift to the right and AgCl_(s) dissolves.

Titrations and Maximum Ion Concentration Worksheet # 9

1. In a titration 25.0ml of a 0.250M AgNO₃ solution was used to precipitate out all of the Br^- in a 200 ml sample. Calculate [Br^-].

Ag⁺ + Br^- → AgBr_(s)

0.0250 L 0.200 L

0.250 M ? M

0.0250 L x 0.250 mole x 1 mole Br^-

[Br^-] = 1L 1 mole Ag⁺

0.200 L

[Br^-] = 0.0313M

2. In a titration 26.5ml of .100M Pb(NO₃)₂ was used to precipitate out all of the Cl^- in a 30.0 ml sample of water. Calculate [Cl^-].

[Cl^-] = 0.177 M

Maximum Ion Concentration

3. Calculate the maximum concentration of OH^- that can exist in a 0.200 M Mg(N0₃)₂ solution.

Mg(OH)_2(s) ⇄ Mg²⁺ + 2OH^-

0.200 M [OH^-]

Ksp = [Mg²⁺][OH^-]²

5.6 x 10^-12 = [0.200][OH^-]²

[OH^-] = 5.3 x 10^-6 M

4. Calculate the maximum concentration of CO₃^-2 that can exist in a .500M AgNO₃ solution.

[CO₃^-2] = 3.4 x 10^-11 M

5. Calculate the maximum concentration of IO₃^- that can exist in a .200M Cu(N0₃)₂ solution.

[IO₃^-] = 5.9 x 10^-4 M

6. Calculate the maximum concentration of Ca⁺² that can exist in a .200M Na₂C0₃ solution.

[Ca²⁺] = 2.5 x 10^-8 M

7. Calculate the minimum number of moles of Pb(NO₃)₂ required to start precipitation in 50.0 mL of 0.15 M ZnCl₂.

PbCl₂ ⇄ Pb²⁺ + 2Cl^-

[Pb²⁺] 0.30 M

Ksp = [Pb²⁺][ Cl^-]²

Don’t forget to multiple 0.15 M by 2 due to ZnCl₂

1.2 x 10^-5 = [Pb²⁺][ 0.30]²

[Pb²⁺] = 1.33 x 10^-4 M

0.0500 L x .000133 moles = 6.7 x 10^-6 moles

L

8. In a titration 12.5 mL of 2.00 x 10^-5 M HCl is required to neutralize 250 mL of saturated AgOH solution. Calculate the [OH^-] and then determine the Ksp for AgOH.

Ag⁺ + Cl^- → AgCl_(s)

.250 L .0125 L

? M .00002 M

[Ag⁺] = 0.0125 L Cl^- x 0.00002 moles x 1 mole Ag⁺

L 1 mole Cl^-

0.250 L

= 1.0 x 10^-6 M

AgOH_(s) ⇄ Ag⁺ + OH^-

1.0 x 10^-6 M 1.0 x 10^-6 M 1.0 x 10^-6 M

Ksp = [Ag⁺][OH^-]

Ksp = (1.0 x 10^-6 )²

Ksp = 1.0 x 10^-12

9. When excess Ag₂CO_3(s) is shaken with 1.00 L of 0.200 M K₂CO₃ it is determined that 6.00 x 10^-6 moles of Ag₂CO₃ dissolves. Calculate the solubility product of Ag₂CO₃.

Ag₂CO_3(s) ⇄ 2Ag⁺ + CO₃^2-

6.00 x 10^-6 M x 2 1.20 x 10^-5M 0.200 M This molarity is determined by the soluble K₂CO₃

There is a 2 in the formula Ag₂CO₃

Ksp = [Ag⁺]²[CO₃^2-]

= [1.20 x 10^-5]²[0.200]

= 2.88 x 10^-11