SOLUBILITY and PRECIPITATION

Topics & Terms

- Solubility
- Molarity
- Limiting Reactant
- Common-Ion Effect
- Solubility Product (Ksp)
- Precipitation

**Helpful Hints....**

- Write the balanced ionic equation for the reaction.
Determine initial concentrations
**moles per liter**of reactants and products. (I = initial concentrations) - Express the changes that will occur as the reaction proceeds. Subtract molarity concentration on the reactants side and add molarity concentration on the product side. (C = change concentrations)
- Express the equilibrium concentrations by combining the change and the initial concentrations. (E = equilibrium concentrations)
- Write the equilibrium expression, substitute concentrations and solve for the unknown.

**Problem 1**

The value of Ksp of AgCl is 1.8 x 10^{-10}. What would be the molar concentration of Ag^{+} and Cl^{-} in AgCl in pure water placed in contact with solid AgCl(s)?

(a) 1.8 x 10 ^{-10} M

(b) 1.3 x 10^{ -5} M

(c) 3.24 x 10^{ -20} M

**Problem 2**

What are the molar concentrations of ions in solution when solid PbI_{2} is in contact with pure water? (Ksp for PbI_{2} = 7.9 x 10^{-9})

(a) [Pb^{+2 }] = 1.3 x 10^{-3 } M & [I^{-}] = 2.6 x 10^{ -3} M

(b) [Pb^{+2}] = 1.6 x 10^{-3} M & [I^{-}] = 3.2 x 10^{-3} M

(c) [Pb^{+2}] = [I^{-}] = 2.0 x 10^{-3} M

**Problem 3**

What is the molar concentration of [Ag^{+}] in AgCl solution in 0.10 M NaCl ?

(Ksp = 1.8 x 10^{-10}) Remember that in this case the molar solubility of AgCl is equal to the [Ag^{+}] as only the Ag^{+} reflects the amount of AgCl that dissolved.

(a) Ag+ = 1.3 x 10^{-5} M

(b) Ag+ = 0.10 M

(c) Ag+ = [AgCl] = 1.8 x 10^{-9} M

**Problem 4**

The molar solubility of PbCl_{2} in 0.10 M NaCl is 1.7 x 10^{-3} moles in a liter (that is 1.7 x 10^{-3} moles of PbCl_{2} will dissolve in 1 liter of the solution). What is the Ksp of PbCl_{2}?

(a) Ksp = 1.8 x 10^{-5}

(b) Ksp = 1.9 x 10^{-8}

(c) Ksp = 1.7 x 10^{-4}

**Problem 5**

Will CaSO_{4} precipitate if 50 mL of 0.0010 M CaCl_{2} is added to 50 mL of 0.010 M Na_{2}SO_{4}?

(Ksp for CaSO_{4} = 2.4 x 10^{-5})

(a) no precipitation will occur

(b) it is at equilibrium

(c) precipitation will occur

**Problem 6**

How many grams (g) of CaSO_{4} is formed when 20 mL of 0.010 M Na_{2}SO_{4} is added to 100 mL of 0.001 M CaCl_{2}?

(a) 0.0001g

(b) 0.01g

(c) 0.0002g