Solubility

Review

1. Solubility is a measure of the maximum amount of solid that will dissolve in a volume of water.

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Units: g/L mol/L g/100mL

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Or even: mL/L for CO_2(g) in water

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Ksp

Solubility Product

Saturated solutions

No Units

Only Changes with Temperature

No ICE!

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Step #1 Write a dissociation equation for the low solubility salt

2. The solubility (s) of BaCO₃ is 5.1 x 10^-5 M @ 25⁰ C. Calculate the solubility product or Ksp.

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Dissociation net ionic BaCO_3(s) ⇌ Ba²⁺ + CO₃^2-

s s s

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Ksp = [Ba²⁺][CO₃^2-]

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Ksp = s²

Ksp = (5.1 x 10^-5)²

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Ksp = 2.6 x 10^-9

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Adding BaCO₃ does not increase [Ba²⁺] or [CO₃^2-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic: Ba²⁺ + CO₃^2- → BaCO_3(s)

2. The solubility (s) of BaCO₃ is 5.1 x 10^-5 M @ 25⁰ C. Calculate the solubility product or Ksp.

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BaCO_3(s) ⇌ Ba²⁺ + CO₃^2-

s s s

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Ksp = [Ba²⁺][CO₃^2-]

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Ksp = s²

Ksp = (5.1 x 10^-5)²

Ksp = 2.6 x 10^-9

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Adding BaCO₃ does not increase [Ba²⁺] or [CO₃^2-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic: Ba²⁺ + CO₃^2- → BaCO_3(s)

Ba²⁺

CO₃^2-

BaCO_3(s)

3. PbCl_2(s) ⇌ Pb²⁺ + 2Cl^-

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Ksp = 4s³

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4. 200.0 mL 0.10 M Pb(NO₃)₂ is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur?

PbCl_2(s) ⇌ Pb²⁺ + 2Cl^-

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200 0.10 M 300 0.20 M 500 500

0.040 M 0.12 M

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TIP = [Pb²⁺][Cl^-]²

TIP = [0.040][0.12] ²

= 5.8 x 10^-4

Ksp = 1.2 x 10^-5 TIP > Ksp ppt forms

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5. Changing the Solubility of a Salt

E + Ca(OH)_2(s) ⇄ Ca²⁺ + 2OH^-

Substance Added Effect on Molar Solubility Effect on Ksp

Ca(NO₃)₂ Decrease none

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Na₂SO₄ Increase none

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NaOH Decrease none

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HCl Increase none

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NaCl none none

Ca(OH)₂ none none

Increase Temperature Increase Increase

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Decrease Temperature Decrease Decrease

6. Mg(OH)₂ will have the greatest solubility in:

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Mg(OH)_2(s) ⇌ Mg²⁺ + 2OH^-

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A. NaOH OH^- lowers solubility

B. Mg(NO₃)₂ Mg²⁺ lowers solubility

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C. H₂O No effect solubility

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D. AgNO₃ Ag⁺ increases solubility by reacting with OH^-

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7. Calculate the maximum number of grams BaCl₂ that will dissolve in 0.50 L of 0.20 M AgNO₃ solution.

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AgCl_(s) ⇄ Ag⁺ + Cl^-

0.20 M

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Ksp = [Ag⁺][Cl^-]

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1.8 x 10^-10 = [0.20][Cl^-]

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[Cl^-] = 9.0 x 10^-10 M

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BaCl_2(s) ⇄ Ba²⁺ + 2Cl^-

4.5 x 10^-10 M 9.0 x 10^-10 M

0.50 L x 4.5 x 10^-10 mole x 208.3 g = 4.7 x 10^-8 g

1 L mole

8. Ionic Solutions: NaCl HCl NH₄NO₃

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Molecular Solutions: C₁₂H₂₂O₁₁

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9. What is the answer?

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A. NaCl

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B. CaCl₂

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C. AgCl

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D. AgBr

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10. Ionic Solutions: NaCl HCl NH₄NO₃

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Molecular Solutions: C₁₂H₂₂O₁₁

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11. What is the answer?

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A. NaCl High

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B. CaCl₂ High

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C. AgCl Low 1.8 x 10^-12

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D. AgBr Low 5.4 x 10^-13

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