UNIT III – REVIEW
1. All ionic compounds (salts) are soluble in water to some extent. Each solid has its
own unique solubility.
Some solids have high solubility, some have low solubility.
Solubility is defined as the concentration of a saturated solution of a solute in a
solvent, at a given temperature.
A saturated solution is one in which dissolved ions are in equilibrium with
undissolved solute.
2. Solubility tables can indicate whether salts are high solubility or low solubility.
Alkali metal (Group I) compounds }
Ammonium compounds
} ALL SOLUBLE
Nitrate compounds
}
3. Low solubility salts establish an equilibrium in solution.
SOLID  nCATIONS + mANIONS
e.g. PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
4. Balanced, net ionic equations show the equilibrium between ions in solution and
undissolved solid.
5. Concentration of ions in solution:
 When solutions are mixed, dilution occurs; diluted concentrations must be
calculated.
 When some solutions are mixed a precipitation reaction occurs.
6. Since an equilibrium is established, an equilibrium constant can be written.
Keq = [cations]n[anions]m = Ksp
7. Quantitative aspects of Ksp:
 Use of solubility to calculate Ksp
 Use of Ksp to calculate solubility
 Use of Ksp to calculate concentration of other ion
 Use of Trial Ion Product to predict precipitation.
8. Common Ion Effect: A common ion reduces the solubility of a salt.
9. Removing Pollutants and Hardness of Water by precipitation:


Heavy metal ions can be precipitated from water by forming precipitates with
very low Ksp
Temporary hardness can be removed by boiling.
Permanent hardness can only be removed by precipitating with a chemical water
softener such as Na2CO3.
REVIEW QUESTIONS
1. Write dissociation equations and calculate the molar concentration of the
cations and the anions in each of the following solutions:
a. 2.24 mol/L Na2S
b. 0.44 mol/L Fe(NO3)2
c. 0.175 mol/L K3PO4
2. What initial volume of concentrated laboratory hydrochloric acid should be
diluted to prepare 5.00L of 0.125 mol/L solution for an experiment?
3. If water is added to a 25.0 mL sample of 2.70 g/L NaOH until the volume
becomes 4.00L, find the concentration of the final solution.
4. A 25.0 mL sample of a saturated potassium chlorate solution is evaporated to
form 2.16 g of crystals. What is the solubility of potassium chlorate in
g/100mL?
5. A litre of a solution saturated at 25C with calcium oxalate is evaporated to
dryness, giving a 0.0061 g residue. Calculate the solubility product constant
for this salt at 25C.
6. By experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide dissolves in 1L
of aqueous solution at 25C. What is the solubility product constant at this
temperature?
7. The mineral fluorite is calcium fluoride. Calculate the solubility in grams per
litre of calcium fluoride in water for the known solubility product constant
3.4 x 10-11.
8. The concentration of calcium ion in blood plasma is 0.0025M. If the
concentration of oxalate ion is 1.0 x 10-7M, do you expect calcium oxalate to
precipitate? Ksp for calcium oxalate is 2.3 x 10-9.
9. Is a precipitate expected to form at equilibrium when 50.0 mL of 0.0010M
BaCl2 is added to 50.0 mL of 0.00010 M Na2SO4? The solubility product
constant for barium sulphate is 1.1 x 10-10.
10. A solution of 0.00016 M lead(II) nitrate was poured into 456 mL of 0.00023
M sodium sulphate. Would a precipitate of lead(II) sulphate be expected to
form if 255 mL of the lead nitrate solution were added?
ANSWERS
1. a.
Na2S(s)  2Na+(aq) + S2-(aq)
4.48 M
2.24 M
b.
Fe(NO3)2(s)  Fe2+(aq) + 2NO3-(aq)
0.44 M
0.88 M
c.
K3PO4(s)  3K+(aq) + PO43-(aq)
0.525 M
0.175 M
2. 52.1 mL
3. 16.9 mg/L
4. 8.64 g/100 mL
5. 2.3 x 10-9
6. 6.9 x 10-9
7. 1.6 x 10-2 g CaF2/L
8. TIP = 2.5 x 10-10 ; TIP  Ksp no ppt.
9. TIP = 2.5 x 10-8 ; TIP  Ksp  ppt.
10. [Pb2+] = 0.00016 M x 255 mL/711 mL
[SO42-] = 0.00023 M x 456 mL/ 711 mL
TIP = (0.00016 M x 255 mL / 711 mL) x (0.00023 M x 456 mL / 711 mL)