Solubilty Scholarship Answers

1. BaSO_4(s) ⇌ Ba²⁺ + CO₃^2-

Adding Ba(NO₃)₂ increases the [Ba²⁺] which causes the equilibrium to shift to the left and BaSO_4(s) precipitates.

2. MgCl₂ + Na₂CO₃ → MgCO_3(s) + 2NaCl

0.0600 L 0.0400 L

? M 0.0030 M

[MgCl₂] = 0.0400 L Na₂CO₃ x 0.0030 mole x 1 mole MgCl₂

1 L 1 mole Na₂CO₃

0.0600 L

= 0.00200 M

This is a titration problem with a double replacement reaction. It is just like the Acid/Base titration questions from last year!

3. Write a dissociation equation with the compound with low solubility!

Mg(OH)_2(s) ⇌ Mg²⁺ + 2OH^-

x x 2x

Ksp = [Mg²⁺][OH^-]²

5.6 x 10^-12 = 4x³

x = 1.1 x 10^-4 M

[OH^-] = 2x = 2.2 x 10^-4 M Don’t forget the units, sig figs, and that [OH^-] is defined as 2x.

4. Write a dissociation equation with the compound with low solubility!

BaSO_4(s) ⇌ Ba²⁺ + SO₄^2- This is a TIP question!

75 x 0.020 M 125 x 0.01 M

200 200

0.0075 M .00625 M

TIP = [Ba²⁺][SO₄^2-]

= (0.0075)(0.00625)

= 4.7 x 10^-5

Ksp = 1.1 x 10^-10

TIP > Ksp Therefore a precipitate will form.

5. The positive ion must be soluble with all negative ions. It could be H⁺, NH₄⁺, or any alkali ion.

6. Write a dissociation equation with the compound with low solubility!

Get a Ksp from experiment 1 or 2 and use it for experiment 3.

BaF_2(s) ⇌ Ba²⁺ + 2F^-

0.036 M 0.026 M

Ksp = [Ba²⁺][F^-]²

Ksp = (0.036)(0.026)²

= 2.4336 x 10^-5

BaF_2(s) ⇌ Ba²⁺ + 2F^-

0.050 M ? M

Ksp = [Ba²⁺][F^-]²

2.4336 x 10^-5 = (0.050)[F^-]²

[F^-] = 0.022 M Do not forget the sig figs!!

7. Write a dissociation equation with the compound with low solubility! Hello...............

AgCl_(s) ⇌ Ag⁺ + Cl^-

Adding NaCl increases the [Cl^-] and causes the equilibrium to shift to the left and AgCl_(s) precipitates.

8. Ba²⁺ + CO₃^2- → BaCO_3(s) Note this shows the formation of BaCO₃ so it is on the right!

BaCO_3(s) ⇌ Ba²⁺ + CO₃^2- Note this shows the equilibrium for BaCO₃ so it is on the left!

If you look on Acid/Base chart page 6, you will see that CO₃^2- is a base. HNO₃ is an acid and will react with the base CO₃^2- lowering its concentration and causing the equilibrium to shift to the right and the BaCO_3(s) dissolves.

The acid according to the above reaction dissolves the precipitate.

9. Write a dissociation equation with the compound with low solubility! If there are two, write two.

AgCl_(s) ⇌ Ag⁺ + Cl^- AgCO_3(s) + 2Ag⁺ + CO₃^2-

0.1 M 0.1 M

Ksp = [Ag⁺][Cl^-] Ksp = [Ag⁺]²[CO₃^2-]

1.8 x 10^-10 = [Ag⁺][0.1] 6.2 x 10^-12 = [Ag⁺]²[0.1 M]

[Ag⁺] = 1.8 x 10^-9 M [Ag⁺] = 7.9 x 10^-6 M

The white precipitate AgCl will form first because it has the lowest [Ag⁺] maximum. The AgCl_(s) will form when the [Ag⁺] is 1.8 x 10^-9 M whereas the AgCO_3(s) will not form until the [Ag⁺] is 7.9 x 10^-6 M.

10. This is a titration and a Ksp question. Start with the titration part.

Ag⁺ + SCN^- → AgSCN_(s)

0.02500 L 0.01515 L

? M 0.102 M

[Ag⁺] = 0.01515 L SCN^- x 0.102 mole x 1 mole Ag⁺

1 L 1 mole SCN^-

0.02500 L

= 0.061812 M Note: this is the [Ag⁺], so do not mess with it by multiplying it by 2!

Now do the Ksp Part of the question.

Ag₂SO_4(s) ⇌ 2Ag⁺ + SO₄^2-

0.061812 M 0.030906 M

Do not multiply the 0.061812 by 2. You only do that when you have the molar solubility of Ag₂SO₄ or when you are going from here to [Ag⁺]. When you go from [Ag⁺] to [SO₄^2-] you must divide by 2.

Ksp = [Ag⁺]²[SO₄^2-]

= (0.061812)²(0.030906)

= 1.18 x 10^-4

11. a) AgCl

AgI

b) Read the question! It says formation of the more soluble precipitate. The more soluble on is the one that dissolves to make way for the second precipitate, which was less soluble!

Ag⁺ + Cl^- → AgCl_(s) Hello........................it says formation!

12. Write a dissociation equation with the compound with low solubility! Hello.......................

Ag₂CrO_4(s) ⇌ 2Ag⁺ + CrO₄^2-

x 2x x

Ksp = [Ag⁺]²[CrO₄^2-]

1.4 x 10^-17 = 4x³

x = 1.518 x 10^-6 M

Do not forget the Grade 11 stuff!

4.0 L x 1.518 x 10^-6 moles x 331.8 g = 0.0020 g

1 L 1 mole