4.4 Solution Concentration and
Stoichiometry
Solution Key Terms
What type of mixture is also considered a solution? Give an example.
– A homogeneous mixture. An example would be a saline solution,
NaCl(aq).
If I dissolve 2.5g NaCl in 25ml H20, which substance is the solvent and
which is the solute?
– In an aqueous solution, water acts as the solvent (what is doing the
dissolving, and sodium chloride acts as the solute (what is being
dissolved).
• When two different liquids with similar structures are present, the majority
component is the solvent and the minority component is the solute.
Solution Concentration
If I have two sodium hydroxide solutions, we’ll label
them A and B, and A was made by dissolving 1.0 g
NaOH in 50 mL of H2O, and B was made by dissolving
2.5 g NaOH in 50 mL H2O, which would be considered
dilute and which would be considered concentrated?
(A) would be the dilute solution, and (B) would be
considered concentrated. This is due to the fact that
solution (A) was made using less solute relative to
solvent.
Solution Concentration (continued)
• Molarity is the most common way to express
solution concentration.
mol solute
Molarity(M) = --------------L solution  (not L solvent)
Let’s Try a Sample Problem
Calculate the molarity of a solution made by adding 45.4 g of NaNO3 to
a flask and dissolving it with water to create a total volume of 2.50 L.
Step 1: convert g solute to mol solute
1.00 mol NaNO3
45.4 g NaNO3 X ---------------------- = 5.34x10-1 mol NaNO3
85.0 g NaNO3
Step 2: calculate concentration
5.34x10-1 mol NaNO3(s)
M = --------------------------------- = 2.14X10-1 M NaNO3(aq)
2.50 L NaNO3(aq)
Using Molarity as a Conversion Factor
• A known molarity can also be used as a conversion factor between moles of solute
and L of solution.
Sample Problem 2:
How many grams of sucrose (C12H22O11) are in 1.55 L of 0.758 M sucrose solution?
Step 1: calculate moles of solute
Moles solute = M x L = 0.758 mol/L X 1.55 L = 1.17 mol C12H22O11
Step 2: covert mol solute to g solute
342.34 g C12H22O11
1.17 mol C12H22O11 X --------------------------- = 401 g C12H22O11
1.00 mol C12H22O11
Solution Dilution
• The easiest was to dilute a stock solution
(concentrated solution), would be to use the
dilution equation:
M1V1 = M2V2
(M1 and V1 are the concentrated molarity and volume)
(M2 and V2 are the dilute molarity and volume)
F.Y.I.: When a solution is diluted, the number of solute particles remain the same
since M(solution) V(solution) = mol solute.
Let’s Try a Practice Problem
To what volume (in mL) should you dilute 100.0 mL of
a 5.00 M CaCl2 solution to obtain a 0.750 M CaCl2
solution?
(5.00 M)(100.0 mL) = (0.750M)(V2)
(5.00 M)(100.0 mL)
V2 = ------------------------- = 667 mL
0.750 M
Solution Stoichiometry
• A balanced chemical equation can be used to make
conversions between solution volumes and the
amount of solute in moles using the molarities of
the solutions.
To do this type of problem, we usually use the
following sequence of steps:
V(solution A)  mol(solute in A)  mol(solute in B)  V(solution B)
Let’s Try a Practice Problem
What volume (in mL) of a 0.150 M HNO3 solution will completely react
with 35.7 mL of a 0.108 M Na2CO3 solution according to the following
equation?
Na2CO3(aq) + 2HNO3(aq)  2NaNO3 + CO2(g) + H2O(l)
Step 1: calculate moles of Na2CO3
mol Na2CO3 = (0.108 M)(0.0357 L) = 3.86X10-3 mol Na2CO3
Step 2: Set up a mol:mol ratio
3.86X10-3
2 mol HNO3
mol Na2CO3 X -------------------- = 7.72X10-3 mol HNO3
1 mol Na2CO3
Step 3: calculate volume of HNO3 solution
mol HNO3
7.72X10-3 mol HNO3
L HNO3 = ----------------- = ---------------------------- = 5.15X10-2 L HNO3 solution = 51.5 mL HNO3 solution
M HNO3
0.150 M HNO3
Chapter 4 pg. 188 #’s 54, 56, 58 & 60 (just the a’s)
Read 4.5-4.7 pgs. 158-168