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Chapter 9
Topic 3
Chemical
Reactions in Aqueous Solutions
Solutions
General Properties of Aqueous Solutions
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A solution is a homogenous mixture of two or more substances.
The substance present in the largest amount (moles) is referred to as the solvent.
The other substances present are called the solutes.
A substance that dissolves in a particular solvent is said to be soluble in that solvent.
A solution
In a solution
The solute can’t be filtered out.
The solute always stays mixed.
Particles are always in motion.
Volumes may not be additive.
A solution will have different properties than the solvent
Solution is a Homogeneous mixture
Solute is the substance in a smaller quantity
Solvent is the substance in a larger quantity
Concentrated means a lot of solute
Dilute means a little solute is present
soluble versus insoluble substance dissolves or does not dissolve.
Concentration – The amount of solute present in the solution
Solubility, saturation and density do not describe Concentration.
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Concentration of solutions
We need a way to tell how much solute �is in a solution - concentration.
There are many systems - we will cover four.
Weight / volume percent
Volume / volume percent
Weight / weight percent
Molarity
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Mass / Volume %
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Mass/Volume % =
Mass solute
Total Volume
x 100
If 5 grams of NaCl is dissolved in enough water to make 200 mL of solution, what is the mass/volume %?
use g and ml
x 100 = 2.5 m/v%
Saline is a 0.9 m/v% solution of NaCl in water.
5 g
200 ml
Volume %
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Volume % =
Volume Solute
Total Volume
x 100
If 10 mL of alcohol is dissolved in enough water to make 200 mL of solution, what is the concentration?
Use the same units for both
10 mL . x 100 = 5 V%
200 mL
Alcohol in wine is measured as a V%.
Mass % or Weight %
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mass % =
Mass Solute
Total Mass
x 100
If a solution of vinegar contains 0.523 grams of acetic acid in 10.41 g of the vinegar solution, what is the % mass?
Use the same units for both
x 100 = 5.02 mass%
0.523 g of acetic acid
10.41 g of solution
pph, ppt, ppm, ppb
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Calculate the % mass and the ppm of 0.0005 g of lead in 100 g of water.
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Concentration of Solutions
Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.
Other common
rearrangements:
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Concentration of Solutions
Determine the molarity of a 5.00 L solution that contains 235 g of sucrose, C12H22O11.
Solution:
Step 1: Determine the molar mass of sucrose. (342.30 g/mol)
Step 2: Determine the moles of sucrose. (0.68653 moles sucrose)
235 g x = 0.68653 mol
(12 C x 12.01g) + (22 H x 1.008 g) + (11 O x 16.00 g) = 342.30
g
mol
Step 3: Determine the molarity of the solution:
1 mol
342.30 g
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Concentration of Solutions
What mass of sucrose, C12H22O11 is needed to prepare 500.0 mL of a 0.025 M solution?
Solution:
Step 1: Determine the moles of sucrose in the solution.
Step 3: Determine the mass of sucrose.
0.0125 mole x = 4.28 g of sucrose
(12 C x 12.01g) + (22 H x 1.008 g) + (11 O x 16.00 g) = 342.30
g
mol
342.30 g
1 mol
Step 2: Determine the molar mass of sucrose. (342.30 g/mol)
Molarity x Volume of the solution = moles of solute
0.025 M x 0.5000 L = 0.0125 moles
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Concentration of Solutions
How many millilitres of 3.50 M NaOH can be prepared from 75.00 grams of solid NaOH?
Solution:
Step 1: Convert grams to moles. (1.87509 moles NaOH)
Step 2: Use the molarity equation to find liters; convert to mL:
Preparation of the Stock Solution
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Add the weighed solid to the volumetric flask, make sure you rinse the weigh boat to get all the sample into the flask.
Add water to dissolve the sample.
After about ½ of the water volume is added, cap the flask and shake to dissolve all the sample
Continue to add water until the solution volume is to the marker line of known volume.
Cap and shake to mix completely.
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Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from a more concentrated one.
moles of solute before dilution = moles of solute after dilution
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Concentration of Solutions
In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Mc x Vc = Md x Vd
Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution.
(2.00 M CuCl2)(Vc) = (0.100 M CuCl2)(250.0 mL)
Vc = 12.5 mL
To make the solution:
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Concentration of Solutions
What volume of 6.0 M H2SO4 is needed to prepare 500.0 mL of a solution that is 0.25 M H2SO4?
Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution.
Mc x Vc = Md x VLd
Concentration of Solutions
What are the concentrations of ions in a solution that is 0.750 M in barium nitrate?
Solution:
Step 1: Determine if barium nitrate is a strong electrolyte using solubility
rules and, if so, write the equation for the dissociation.
Step 2: Determine the concentration of Ba2+(aq) ions using the
stoichiometric ratio.
Square brackets around a chemical species indicates concentration generally in Molarity.
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Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
Gravimetric analysis is an analytical technique based on the measurement of mass.
Gravimetric analysis is highly accurate.
Applicable only to reactions that go to completion or have nearly 100 % yield.
One common type of gravimetric analysis involves the isolation of a precipitate such as performed in Experiment 9.
Typical steps involve:
1) Mass an unknown solid.
2) Dissolve the unknown in water.
3) React the unknown with an excess amount of a substance that is
known to form a precipitate.
4) Filter, dry and weigh the precipitate.
5) Use the formula mass and the mass of the precipitate to find the %
mass of the unknown ion.
Chapter 4 Unit 1
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Concentration of Solutions
What are the concentrations of ions in a solution that is 0.750 M in barium nitrate? [Ba(NO3)2]
Solution:
Step 3: Determine the concentration of NO3–(aq) ions using the
stoichiometric ratio.
Ba(NO3)2(s)
H2O
Ba2+(aq) + 2NO3(aq)
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Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
A 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample?
Solution:
Step 1: Find the percent by mass chlorine in AgCl.�
Step 2: Multiply the percent of Cl by the mass of the precipitate to obtain the mass of Cl in the sample.
0.247 x 1.725 g AgCl = 0.427 g Cl in the sample
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
A 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample?
Solution:
Step 3: Divide the mass of Cl in sample by the total mass of sample; multiply by 100 to determine percent.
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Aqueous Reactions and Chemical Analysis
Quantitative studies of acid-base neutralization reactions or many redox reactions are most conveniently carried out using a technique known as a titration.
A titration is a volumetric technique that uses volumetric glassware to measure volume precisely, like burets.
If we have an acid base reaction, then the reaction is
Acid + base = salt + water.
The moles of acid will at some point equal the moles of base.
The point in the titration where the acid has been neutralized is called the equivalence point.
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An acid or base solution is usually colorless. So we need a method to visualize this point. We add another compound that is sensitive to the amount of acid or base present or is sensitive the the pH of the solution. The equivalence point can then be signalled by a color change.
The color change is brought about by the use of an indicator.
Indicators have distinctly different colors in acidic and basic media.
The indicator is chosen so that the color change, or endpoint, is very close to the equivalence point.
Phenolphthalein is a common indicator. It changes when the pH is just slightly basic from a clear to a pink color.
Aqueous Reactions and Chemical Analysis
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Aqueous Reactions and Chemical Analysis
Titrations are an analytical technique based on the measurement of volume..
Titrations are highly accurate and precise.
Applicable only to reactions that are in solution involving ionic species such as acids, bases and oxidation / reduction reaction of aqueous ions.
Typical steps of a titration include:
1) Prepare an unknown solution
2) Add an appropriate indicator to visualize the endpoint.
3) Carefully add a standard solution with a buret until the endpoint is
reached.
4) Using solution stoichiometry, calculate the molarity or % mass of
the unknown solution.
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Sodium hydroxide solutions are commonly used in titrations.
NaOH solutions must be standardized because the concentration of the solution changes over time. (NaOH reacts with CO2 that slowly dissolves into the solution forming carbonic acid.) NaOH also cannot be weighed precisely because it absorbs water from the atmosphere very quickly.
The acid, potassium hydrogen phthalate (KHP), is frequently used to standardize NaOH solutions. It is called a primary standard because it can be weighed precisely. It does not absorb water significantly.
acidic proton of KHP;
KHP is a monoprotic acid
KHC8H4O4
Aqueous Reactions and Chemical Analysis
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 1: Use the molar mass of KHP to determine the moles of KHP.
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Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 2: Using the balanced reaction, convert to moles of NaOH
KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l)
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 3: Use the molarity equation to calculate molarity of the sodium
hydroxide solution.
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
How many milliliters of a 1.42 M H2SO4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution?
Solution:
Step 1: Write and balance the chemical equations that corresponds to the
neutralization reaction:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
Step 2: Use the molarity and volume given to determine the moles of
KOH.
moles of KOH = 0.336 M KOH x 0.0955 L = 0.032088 moles KOH
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
How many milliliters of a 1.42 M H2SO4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution?
Solution:
Step 3: Using the moles of KOH and the stoichiometric ratios from the balanced equation, convert to moles of H2SO4.
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
Chapter 4 Unit 1
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Aqueous Reactions and Chemical Analysis
How many millilitres of a 1.42 M H2SO4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution?
Solution:
Step 4: Using the moles of H2SO4 and the concentration given, calculate the millilitres of solution.
Stoichiometry
molB x = massB
massA x = molA
MolarityA x VolumeA = molA
molA x = molB
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1 mol
grams
2 mol of B
1 mol of A
grams
1 mol
1 A + 2 B → 2 C
molB = MB VB