Solution Chemistry (Chp. 7) Chemistry 2202. Solutions Terms Molar Concentration (mol/L) Dilutions %...

Solution Chemistry(Chp. 7)

Chemistry 2202

Solutions Terms Molar Concentration (mol/L) Dilutions % Concentration (pp. 255 – 263) Solution Process Solution Preparation Solution Stoichiometry Dissociation

Termssolution

solvent

solute

concentrated

dilute

aqueous

miscible

immiscible

alloy

solubility

molar solubility

saturated

unsaturated

supersaturated

dissociation

electrolyte

non-electrolyte

soluble

insoluble

limiting reagent

excess reagent

actual yield

theoretical yield

decanting

pipetting

filtrate

precipitate

dynamic equilibrium

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

Terms solution - is a homogeneous mixture solute - the substance that dissolves OR the

substance in lesser quantitysolvent - the substance which dissolves the

solute OR the substance in greater quantity

concentrated - a large amount of solute relative to the amount of solvent

dilute - a small amount of solute relative to the amount of solvent

saturated – contains the maximum amount of dissolved solute at a given temperature and pressure

unsaturated – contains less than the maximum amount of dissolved solute at a given temperature and pressure

supersaturated – contains more than the maximum amount of dissolved solute for a given temperature and pressure

Terms

Termsmiscible – liquids that dissolve in each other

immiscible – liquids that do not dissolve in each other

aqueous - the solvent is water

alloy - a solid solution of two or more metals

Terms

Solubility - the maximum amount of solute that can be dissolved under specific temperature and pressure conditions

eg. the solubility of HCl at 25 °C is 12.4 mol/L

eg. 100.0 mL of water at 25°C dissolves 36.2 g of sodium chloride

soluble – solubility is greater than 1 g per 100 mL of solvent.

insoluble - solubility is less than 0.1 g per 100 mL of solvent.

Terms

9 Types of Solution

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

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.

Solubility small molecules are often more soluble than

larger molecules. solubility of solids increases with temperature. 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.

Factors Affecting Solubility3. What type of solvent will dissolve:

polar solutes and ionic solutes nonpolar solutes

4. Why do some ionic compounds have low solubility in water?

p. 254 #’s 1, 2, 4 - 6

“Like Dissolves Like” ionic solutes and polar covalent solutes

both dissolve in polar solvents non-polar solutes dissolve in non-polar

solvents. compounds with very strong ionic bonds,

such as AgCl, tend to be less soluble in water than compounds with weak ionic bonds, such as NaCl.

Applications

1. An opened soft drink goes ‘flat’ faster if not refrigerated.

2. Thermal pollution (warming lake water) is not 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.

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.

Molar Concentration

The terms concentrated and dilute are qualitative descriptions of solubility.

A quantitative measure of solubility uses numbers to describe how much solute is dissolved or the concentration of a solution.

Molar Concentration

The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).

Molar Concentration

FORMULA:

Molar Concentration = number of moles

volume in litres

C = n

V

eg. Calculate the molar concentration of:1. 4.65 mol of NaOH is dissolved to

prepare 2.83 L of solution. (1.64 mol/L)

2. 15.50 g of NaOH is dissolved to prepare 475 mL of solution.

( 0.3875 mol → 0.816 mol/L)

p. 268 - # 19

eg. Calculate the following:a) the number of moles in 4.68 L of 0.100

mol/L KCl solution. (0.468 mol)

b) the mass of KCl in 268 mL of 2.50 mol/L KCl solution. (0.670 mol → 49.9 g)

p. 268 #’s 20 - 24

C

n V orVxCn Rearranged

Formulas

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.

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

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.

p. 273 #’s 25 – 27

p. 276 #’s 1, 2, 4, & 5

DON’T SHOW UP UNLESS THIS IS DONE!!

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

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

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.

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

Percent Concentration

100x(mL) solutionofvolume

(g) soluteofmass(m/v)Percent

p. 258 #’s 1 – 3 DSUUTID!!

p. 261 #’s 5 – 9

DSUUTID!!

100x(g) solutionofmass

(g) soluteofmass(m/m)Percent

p. 263 #’s 10 – 13

DSUUTID!!

100x(mL) solutionofvolume

(mL) soluteofvolume(v/v)Percent

Concentration in ppm and ppbParts per million (ppm) and parts per

billion (ppb) are used for extremely small concentrations

610solution

solute

mxppmm

910solution

solute

mxppbm

eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in:

a) ppm

b) ppb

ppm = 0.005 g x 106

100,000 g

= 0.05 ppm

ppb = 0.005 g x 109

100,000 g

= 50.0 ppb

p. 265 #’s 15 – 17

pp. 277, 278 #’s 11, 13, 15 – 18, 20

DON’T SHOW UP UNLESS THIS IS DONE!!

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

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

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)?

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

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

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

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)

Ca(NO3)2(s) → Ca2+(aq) + 2 NO3

-(aq)

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. Acids form ions when dissolved in

water.

eg. H2SO4(aq) → 2 H+(aq) + SO4

2-(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.

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)

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