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
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Amount in moles, n = m / M
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Reactions in Solution
• Many reactions occur in solution.
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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.
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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
C  n
V
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Reactions in Solution
n  m C  n
M
V
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
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Reactions in Solution
n  m C  n
M
V
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 = 23.0 + 16.0 + 35.5
= 74.5 g mol-1
n = m/M = 3.00 g / 74.5 g mol-1 = 0.0403 mol
C = n/V
= 0.0403 mol / 0.1 L
= 0.403 M
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C  n
V
Reactions in Solution
Q: What is the concentration of NaOH solution if a 25.00
mL portion of it reacts exactly with 22.40 mL of 0.104
M hydrochloric acid?
A: NaOH + HCl  NaCl + H2O
V (NaOH) = 25.00 mL = 0.02500 L
V (HCl) = 22.40 mL = 0.02240 L
[HCl]
= 0.104 M
n (HCl used) = V x C
= 0.02240 L  0.104 mol L-1
= 0.002330 mol
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Reactions in Solution
C  n
V
A: NaOH + HCl  NaCl + H2O
1 mol of HCl reacts with 1 mol of NaOH
\ 0.002330 mol of HCl reacts with 0.002330 mol of NaOH
\ n (NaOH) = 0.002330 mol
V (NaOH) = 25.00 mL = 0.02500 L
\ [NaOH]
= n/V
= 0.002330 mol / 0.02500 L
= 0.09318 M
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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…)
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Questions to complete for next lecture:
1. Consider dissolving 23.4 g of sodium sulfate (Na2SO4) in
enough water to form 125 mL of solution.
a) Write the equation representing the dissolving of sodium sulfate.
b) What is the molar mass of sodium sulfate?
c) How many moles are there in 23.4 g of sodium sulfate?
d) What is [Na2SO4]?
e) What is [Na+] in this solution?
f)
What is [SO42-] in this solution?
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