1. Recap – Stoichiometry Balanced chemical equation Convert mass to moles
Chem 1001 Lecture 11
Recap – Stoichiometry
Balanced chemical equation
Convert mass to moles
Check for limiting reagent
Use moles and equation coefficients to obtain amount of target needed
Covert moles of target to mass
Calculate % yield
Amount in moles, n = m / M

2. Reactions in Solution Many reactions occur in solution.
Chem 1001 Lecture 14
Reactions in Solution
Many reactions occur in solution.
Dr Toby Hudson

3. Chem 1001 Lecture 14
Reactions in Solution
The solvent is the substance that does the dissolving.
The solute is the substance that is dissolved.
The solution is the mixture.
Concentration refers to amount of solute dissolved in 1 L of solution.
Dr Toby Hudson

4. Chem 1001 Lecture 14
Reactions in Solution
In chemistry we use: no moles solute dissolved in 1000 mL (1 L) of solution - Molarity (units: molar = M).
So a 1 molar (1 M) solution contains 1 mole of solute in 1 L of solution.
Concentration in M = no. moles / Vol in L
Dr Toby Hudson

5. Chem 1001 Lecture 14
Reactions in Solution
Q: A saline solution used in pharmaceuticals contains 0.10 M sodium chloride. What mass is needed to prepare 10 L of saline?
A: C = 0.1 M, V = 10. L
n = CV
= 0.10 mol L-1  10. L
= 1.0 mol
m = n  M
= 1.0 mol  58.5 g mol-1
= 58.5 g
Dr Toby Hudson

6. Chem 1001 Lecture 14
Reactions in Solution
Q: A 3% w/v bleach solution contains 3.00 g sodium hypochlorite (NaOCl) in 100 mL of solution. What is its molarity?
A: Molar mass of NaOCl = = 74.5 g mol-1
n = m/M = 3.00 g / 74.5 g mol-1 = mol
C = n/V = mol / 0.1 L = M
Dr Toby Hudson

7. Chem 1001 Lecture 14
Reactions in Solution
Q: What is the concentration of NaOH solution if a mL portion of it reacts exactly with mL of M hydrochloric acid?
A: NaOH + HCl  NaCl + H2O
V (NaOH) = mL = L
V (HCl) = mL = L
[HCl] = M
n (HCl used) = V x C
= L  mol L-1
= mol
Dr Toby Hudson

8. Reactions in Solution A: NaOH + HCl  NaCl + H2O
Chem 1001 Lecture 14
Reactions in Solution
A: NaOH + HCl  NaCl + H2O
  1 mol of HCl reacts with 1 mol of NaOH  
\ mol of HCl reacts with mol of NaOH
\ n (NaOH) = mol
V (NaOH) = mL = L
\ [NaOH] = n/V
= mol / L
= M
Dr Toby Hudson

9. Learning Outcomes: By the end of this lecture, you should:
know the meaning of the terms: solute, solvent and solution
understand that molarity represents a concentration in moles per litre of solution
be able to calculate concentrations given the volume and number of moles or mass of substance
be able to use concentration and volume to determine amount of substance present in moles
be able to complete the worksheet (if you haven’t already done so…)

10. Questions to complete for next lecture:
Chem 1001 Lecture 4
Questions to complete for next lecture:
Consider dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution.
Write the equation representing the dissolving of sodium sulfate.
What is the molar mass of sodium sulfate?
How many moles are there in 23.4 g of sodium sulfate?
What is [Na2SO4]?
What is [Na+] in this solution?
What is [SO42-] in this solution?