IB Chemistry on Molarity, Concentration, standard solution serial dilution
Solution Chemistry (Chp. 7) Chemistry 2202. Topics Molar Concentration (mol/L) Dilutions %...
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Transcript of Solution Chemistry (Chp. 7) Chemistry 2202. Topics Molar Concentration (mol/L) Dilutions %...
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Solution Chemistry(Chp. 7)
Chemistry 2202
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Topics Molar Concentration (mol/L) Dilutions % Concentration (pp. 255 – 263) Solution Process Solution Preparation Solution Stoichiometry Dissociation
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Termssolution
solvent
solute
concentrated
dilute
aqueous
miscible
Immiscible
alloy
solubility x
molar solubility
saturated
unsaturated
supersaturated
dissociation
electrolyte
non-electrolyte
filtrate
precipitate
limiting reagent
excess reagent
actual yield
theoretical yield
decanting
pipetting
dynamic equilibrium
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Define the terms in bold and italics from pp. 237 – 240.
Solids, liquids, and gases can combine to produce 9 different types of solution. Give an example of each type.
p. 242 #’s 5, 7, 9, & 10
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Termssolution
solvent
solute
concentrated
dilute
aqueous
miscible
immiscible
x
alloy
solubility
molar solubility
saturated
unsaturated
supersaturated
dynamic equilibrium
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Factors Affecting Solubility (pp.243 – 254)
1. List 3 factors that affect the rate of dissolving.
2. How does each of the following affect solubility?
particle size temperature pressure
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Factors Affecting Solubility3. What type of solvent will dissolve:
polar solutes nonpolar solutes ionic solutes
4. Why do some ionic compounds have low solubility in water?
p. 254 #’s 1, 2, 4 - 6
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Section 7.2 (pp. 243 – 252)
State the generalizations regarding solubility and solutions (in italics)
Define terms (in bold)
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rate of dissolving dipole
ion-dipole attractions
hydrated
electrolyte.
non-electrolytes
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Rate of Dissolving
for most solids, the rate of dissolving is greater at higher temperatures
stirring a mixture or by shaking the container increases the rate of dissolving.
decreasing the size of the particles increases the rate of dissolving.
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“Like Dissolves Like”
ionic solutes and polar covalent solutes both dissolve in polar solvents
non-polar solutes dissolve in non-polar solvents.
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Solubility small molecules are often more
soluble than larger molecules. the solubility of most solids increases
with temperature. the solubility of most liquids is not
greatly affected by temperature. the solubility of gases decreases as
temperature increases
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Solubility
An increase in pressure increases the solubility of a gas in a liquid.
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Applications
1. An opened soft drink goes ‘flat’ faster if not refrigerated.
2. Warming of pond water may not be healthy for the fish living in it.
3. After pouring 5 glasses of pop from a 2 litre container, Jonny stoppered the bottle and crushed it to prevent the remaining pop from going flat.
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Molar Concentration
Review:
- Find the molar mass of Ca(OH)2
- How many moles in 45.67 g of Ca(OH)2?
- Find the mass of 0.987 mol of Ca(OH)2.
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Molar Concentration
The terms concentrated and dilute are qualitative descriptions of solubility.
A quantitative measure of solubility uses numbers to describe the concentration of a solution.
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Molar Concentration
The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).
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Molar Concentration
FORMULA:
Molar Concentration = number of moles
volume in litres
C = n
V
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eg. Calculate the molar concentration of: 4.65 mol of NaOH is dissolved to
prepare 2.83 L of solution.
15.50 g of NaOH is dissolved to prepare 475 mL of solution.
p. 268 - # 19
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Eg. Calculate the following:
a) the number of moles in 4.68 L of 0.100 mol/L KCl solution.
b) the mass of KCl in 268 mL of 2.50 mol/L KCl solution.
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c) the volume of 6.00 mol/L HCl(aq) that can be made using 0.500 mol of HCl.
d) the volume of 1.60 mol/L HCl(aq) that can be made using 20.0 g of HCl.
p. 268 #’s20-24
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Dilution (p. 272)
When a solution is diluted:- The concentration decreases- The volume increases- The number of moles remains
the same
ni = nf Number of moles after dilution
Number of moles before dilution
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Dilution (p. 272) ni = nf
Ci Vi = Cf Vf
eg. Calculate the molar concentration of a vinegar solution prepared by diluting 10.0 mL of a 17.4 mol/L solution to a final volume of 3.50 L.
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p. 273 #’s 25 – 27
p. 276 #’s 1, 2, 4, & 5
DON’T SHOW UP UNLESS THIS IS DONE!!
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Solution Preparation & Dilution standard solution – a solution of known
concentration volumetric flask – a flat-bottomed glass vessel
that is used to prepare a standard solution delivery pipet – pipets that accurately measure
one volume graduated pipet – pipets that have a series of
lines that can be use to measure many different volumes
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To prepare a standard solution:
1. calculate the mass of solute needed
2. weigh out the desired mass
3. dissolve the solute in a beaker using less than the desired volume
4. transfer the solution to a volumetric flask (rinse the beaker into the flask)
5. add water until the bottom of the meniscus is at the etched line
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To dilute a standard solution: 1. Rinse the pipet several times with
deionized water. 2. Rinse the pipet twice with the
standard solution. 3. Use the pipet to transfer the required
volume. 4. Add enough water to bring the
solution to its final volume.
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Percent Concentration Concentration may also be given as a %. The amount of solute is a percentage of
the total volume/mass of solution. liquids in liquids - % v/v solids in liquids - % m/v solids in solids - % m/m
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Percent Concentration
100x(mL) solutionofvolume
(g) soluteofmass(m/v)Percent
p. 258 #’s 1 – 3 DSUUTID!!
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p. 261 #’s 5 – 9
DSUUTID!!
100x(g) solutionofmass
(g) soluteofmass(m/m)Percent
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p. 263 #’s 10 – 13
DSUUTID!!
100x(mL) solutionofvolume
(mL) soluteofvolume(v/v)Percent
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Concentration in ppm and ppbParts per million (ppm) and parts per
billion (ppb) are used for extremely small concentrations
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610solution
solute
mxppmm
910solution
solute
mxppbm
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eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in:
a) ppm
b) ppb
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ppm = 0.005 g x 106
100,000 g
= 0.05 ppm
ppb = 0.005 g x 109
100,000 g
= 50.0 ppb
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p. 265 #’s 15 – 17
pp. 277, 278 #’s 11, 13, 15 – 18, 20
DON’T SHOW UP UNLESS THIS IS DONE!!
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Solution Stoichiometry
1. Write a balanced equation
2. Calculate moles given
n=m/M OR n=CV3. Mole ratios
4. Calculate required quantity
nMmORV
nCOR
C
nV
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Solution Stoichiometry
eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L NaOH solution.
Calculate the molar concentration of the HCl(aq) solution.
p. 304: #’s 16, 17, & 18
Worksheet
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Sample Problems1. What mass of copper metal is needed
to react with 250.0 mL of 0.100 mol/L silver nitrate solution?
2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH.
3. What volume of 3.00 mol/L HNO3(aq) is needed to neutralize 450.0 mL of 0.100 mol/L Sr(OH)2(aq)?
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Sample Problem Solutions
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Step 2n = 0.02500 mol AgNO3
Step 3n = 0.01250 mol Cu
Step 4m = 0.794 g Cu
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Sample Problem Solutions
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2n = 0.0300 mol NaOH
Step 3n = 0.0300 mol HCl
Step 4V = 0.0150 L HCl
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Sample Problem Solutions
2 HNO3(aq) + Sr(OH)2(aq) →
2 H2O(l) + Sr(NO3)2(aq)
Step 2n = 0.04500 mol Sr(OH)2
Step 3n = 0.0900 mol HNO3
Step 4V = 0.0300 mol/L HNO3
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The Solution Process (p. 299)
Dissociation occurs when an ionic compound breaks into ions as it dissolves in water.
A dissociation equation shows what happens to an ionic compound in water.
eg. NaCl(s) → Na+(aq) + Cl-(aq)
K2SO4(s) → 2 K+(aq) + SO4
2-(aq)
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The Solution Process (p. 299) Solutions of ionic compounds conduct
electric current. A solute that conducts an electric
current in an aqueous solution is called an electrolyte.
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The Solution Process (p. 299)
Acids are also electrolytes. Acids form ions when dissolved in
water.
eg. H2SO4(aq) → 2 H+(aq) + SO4
2-(aq)
HCl(s) → H+(aq) + Cl-(aq)
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Molecular Compounds DO NOT dissociate in water.
eg. C12H22O11(s) → C12H22O11(aq)
Because they DO NOT conduct electric current in solution, molecular compounds are non-electrolytes.
The Solution Process (p. 299)
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The molar concentration of any dissolved ion is calculated using the ratio from the dissociation equation.
eq. What is the molar concentration of each ion in a 5.00 mol/L MgCl2(aq) solution:
5.00 mol/L 5.00 mol/L 10.00 mol/L
The Solution Process (p. 299)
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p. 300 #’s 7 – 9p. 300 #’s 7 – 9
What mass of calcium chloride is What mass of calcium chloride is required to prepare 2.00 L of 0.120 required to prepare 2.00 L of 0.120 mol/L Clmol/L Cl--(aq)(aq) solution? solution?
p. 302 # 14p. 302 # 14
p. 311 #’s 11, 12, 16, & 18p. 311 #’s 11, 12, 16, & 18