Solution Stoichiometry – Concentration
of Ions
To complete a question like this, you must:
- Create a balanced equation showing the dissociation of the solid salt into its aqueous ions. Be sure to include state symbols.
- List known and unknown data in columns beneath appropriate substances.
- Determine the number of moles of solid salt.
- Set up a mole ratio between the salt and its ion.
- Determine the concentration of the aqueous ion.
ex.
If 21.6 g of sodium phosphate is dissolved in 125 mL of solution,
determine the molar concentration of sodium ions.
Na3PO4(s) → 3Na+(aq) + PO43-(aq)
m
= 21.6 g
V = 125 mL
M
= 163.94 g/mol c = ?
n = 0.395 mol
n
= m/M
= 21.6
g/163.94 g/mol c = n/V
= 0.132 mol
= 0.395 mol/0.125 L
= 3.16 mol/L
nNa3PO4/1 = nNa+/3
nNa+ = 3(0.132 mol)
= 0.395 mol
Solution Stoichiometry – "Regular"
Complete this question like any other stoichiometry problem we have done in the past. Just be aware that we now have another equation available: n = cV.
ex.
If 29.4 g of potassium is added to 197 mL of a 1.346 mol/L solution of
aluminum bromide,
(a)
what is the concentration of the potassium bromide solution produced?
(b)
what mass of aluminum is produced?
 |
| Please note that I have used an alternate method to determine lim/xs reactant. However, it gives the same results as using the comparison factor (CF). |
Please note that the volume used for the potassium bromide product solution is 0.197 L, which is the same as the volume of the aluminum bromide reactant solution. When carrying out solution stoich calculations, the total volume of aqueous solution with which we start will be the volume that any soluble product will use.
Essentially, we are assuming that any aqueous reactant solution is pure water (which, of course, is not entirely true) because it makes the calculations simpler.
In this case, we start with 197 mL of aqueous reactant, so the soluble product will have 197 mL available in which to dissolve.
If we had a more involved situation like this, we would add the two reactant solution volumes to find the soluble product volume:
A(aq) + B(aq) → C(aq) + D(s)
V = 200 mL V = 150 mL V = 350 mL
Essentially, we are assuming that any aqueous reactant solution is pure water (which, of course, is not entirely true) because it makes the calculations simpler.
In this case, we start with 197 mL of aqueous reactant, so the soluble product will have 197 mL available in which to dissolve.
If we had a more involved situation like this, we would add the two reactant solution volumes to find the soluble product volume:
A(aq) + B(aq) → C(aq) + D(s)
V = 200 mL V = 150 mL V = 350 mL
Homework:
Answers:
Student Questions:
1. For the first homework question, how do you know the volume of the calcium hydroxide?
In this case, some of the water reactant would get used up in the reaction, but we will keep it simple and ignore that to keep the calculation easy. So, we assume that all 800 mL of water remains after the reaction is complete and the the soluble calcium hydroxide product will dissolve in that 800 mL.
