Solution Chemistry(Chp. 7)

Chemistry 2202

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Topics Definitions/properties 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 - 67

Section 7.2 (pp. 243 – 252) State the generalizations regarding

solubility and solutions (in italics) Define terms (in bold)

ion-dipole attractions electrolyte

hydrated non-electrolyte

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 while the solubility of most liquids is not affected by temperature.

the solubility of gases decreases as temperature increases

an increase in pressure increases the solubility of a gas in a liquid.

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

V17

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 #’s20-2420

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

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

OR n=CV3. Mole ratio

4. Calculate required quantity

nMmORV

nCOR

C

nV

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M

mn

Solution Stoichiometry

eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L Ca(OH)2 solution.

Calculate the molar concentration of the HCl(aq) solution.

p. 304: #’s 16, 17, & 18

Worksheet37

Sample Problems1. What mass of copper metal is needed to

react with 250.0 mL of 0.100 mol/L silver nitrate solution? (0.794 g Cu)

Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

Step 2 n = 0.02500 mol AgNO3

Step 3 n = 0.01250 mol Cu

Step 4 m = 0.794 g Cu

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2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH. (0.0150 L HCl)

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Step 2 n = 0.0300 mol NaOH

Step 3 n = 0.0300 mol HCl

Step 4 V = 0.0150 L HCl

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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)? (0.0300 L HNO3)

2 HNO3(aq) + Sr(OH)2(aq) → 2 H2O(l) + Sr(NO3)2(aq)

Step 2 n = 0.04500 mol Sr(OH)2

Step 3 n = 0.0900 mol HNO3

Step 4 V = 0.0300 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.

The Solution Process (p. 299) Acids are also electrolytes.

6 strong acids: perchloric acid - HClO4(aq)

hydrochloric acid – HCl(aq)

hydroiodic acid - HI(aq)

hydrobromic acid – HBr(aq)

nitric acid – HNO3(aq)

sulfuric acid – H2SO4(aq)

eg. H2SO4(aq) → 2 H+(aq) + SO4

2-(aq)

HCl(s) → H+(aq) + Cl-(aq)

All other acids are weak electrolytes(poor conductors)

The Solution Process (p. 299)eg. H2SO4(aq) → 2 H+

(aq) + SO42-

(aq)

HCl(s) → H+(aq) + Cl-(aq)

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.

Low solubility compounds are weak electrolytes due to their low solubility in water.

eg. AgBr BaSO4 PbI2

The Solution Process (p. 299)

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)

eg. Calculate the concentration of Al(NO3)3 in a solution with a nitrate ion concentration equal to 0.300 mol/L.

The Solution Process (p. 299)

eg. Calculate the molar concentration of chloride ions in a solution prepared by dissolving 10.0 g of FeCl3 to make 100.0 mL of solution

The Solution Process (p. 299)

eg. Calculate the mass of Na2SO4 needed to prepare 400.0 mL of solution with a sodium ion concentration equal to 0.350 mol/L.

The Solution Process (p. 299)

eg. What mass of calcium chloride is required to prepare 2.00 L of 0.120 mol/L Cl-(aq) solution?

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