Lecture, No-4 Lecture Objectives

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By the end of this lesson the student is

expected ………..

Lecture, No-4

1.

To define concentration and dilution

terms.

2.

To study the Precision and cost of

glassware

3.

To Prepare solutions from different

materials in different concentration.

Lecture Objectives

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Concentration =

# of fish

volume (L)

Concentration =

V = 1000 mL

n = 2 fish

Concentration = 2 “fishar”

V = 1000 mL

n = 4 fish

[ ] = 4 “fishar”

V = 5000 mL

n = 20 fish

[ ] = 4 “fishar”

Concentration = 1 “fishar”

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V = 1000 mL

n = 2 moles

Concentration = 2 molar

V = 1000 mL

n = 4 moles

[ ] = 4 molar

V = 5000 mL

n = 20 moles

[ ] = 4 molar

Concentration =

# of moles

volume (L)

V = 250 mL n = 8 moles [ ] = 32 molar

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Making

Molar

Solutions

…from liquids

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Concentration

…a measure of solute-to-solvent ratio

concentrated

vs. dilute

“lots of solute”

“not much solute”

“watery”

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Making a

Dilute

Solution

initial solution remove sample diluted solution same number of moles of solute in a larger volume mix moles of solute

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Concentration

“The amount of solute in a solution”

mol L

_M

A. mass % = mass of solute mass of sol’n

B. parts per million (ppm)  also, ppb and ppt

–

commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute

L of sol’n

–

used most often in this class

D. molality (m) = moles of solute kg of solvent

M =

mol

L

% by mass – medicated creams % by volume – rubbing alcohol

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precise; expensive Range:

Glassware – Precision and Cost

beaker vs. volumetric flask

When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = 50 mL volumetric flask 1000 mL + 0.30 mL 950 mL – 1050 mL 999.70 mL– 1000.30 mL imprecise; cheap Range:

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Markings on Glassware

TC 20

o

C “to contain at a temperature of 20

o

C”

TD “to deliver”

T

s

“time in seconds”

22

500 mL + 5%

Range = 500 mL + 25 mL

475 – 525 mL

Beaker

Graduated Cylinder

Volumetric Flask 500 mL + 0.2 mL

Range = 499.8 – 500.2 mL

1000 mL + 5 mL Range = 1000 mL + 5 mL

475 – 525 mL

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~ ~ ~ ~ ~ ~ ~ ~ water in grad. cyl. mercury in grad. cyl.

Measure to part of meniscus w/zero slope.

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How to mix solid chemicals

Lets mix chemicals for the upcoming soap lab.

We will need 1000 mL of 3 M NaOH per class.

How much sodium hydroxide will I need, for five classes, for this lab?

M =

mol

L

3 M =

? mol

1 L

? = 3 mol NaOH/class

How much will this weigh?

1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol

MM

_NaOH

= 40g/mol

40.0 g NaOH 1 mol NaOH X g NaOH = 15.0 mol NaOH =

To mix this, add 120 g NaOH into 1L volumetric flask with

~750 mL cold H

₂

O. Mix, allow to return to room temperature – bring volume to 1 L.

FOR EACH CLASS:

x 5 classes

15 mol NaOH

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How to mix a Standard Solution

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

Wash bottle Volume marker (calibration mark) Weighed amount of solute

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How to mix a Standard Solution

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Process of Making a Standard

Solution from Liquids

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How to mix a dilute solution from a

concentrated stock solution

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Identify each volume to two decimal places

(values tell you how much you have expelled)

4.48 - 4.50 mL

_{4.86 - 4.87 mL}

_{5.00 mL}

Reading a pipette

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Dilution of Solutions

Solution Guide Formula

Weight Specific Gravity Molarity Reagent Percent To Prepare 1 Liter of one molar

Solution

Acetic Acid Glacial (CH₃COOH) 60.05 1.05 17.45 99.8% 57.3 mL Ammonium Hydroxide (NH₄OH) 35.05 0.90 14.53 56.6% 69.0 mL Formic Acid (HCOOH) 46.03 1.20 23.6 90.5% 42.5 mL Hydrochloric Acid (HCl) 36.46 1.19 12.1 37.2% 82.5 mL Hydrofluoric Acid (HF) 20.0 1.18 28.9 49.0% 34.5 mL Nitric Acid (HNO₃) 63.01 1.42 15.9 70.0% 63.0 mL Perchloric Acid 60% (HClO₄) 100.47 1.54 9.1 60.0% 110 mL Perchloric Acid 70% (HClO₄) 100.47 1.67 11.7 70.5% 85.5 mL Phosphoric Acid (H₃PO4) 97.1 1.70 14.8 85.5% 67.5 mL Potassium Hydroxide (KOH) 60.05 1.05 17.45 99.8% 57.3 mL Sodium Hydroxide (NaOH) 40.0 1.54 19.4 45.0% 85.5 mL Sulfuric Acid (H₂SO₄) 98.08 1.84 18.0 50.5% 51.5 mL

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D D C C V M V M  L) (25.00 M 0.500 ) (V M 14.8 V M V M_C _C  _D _D  _C  C = concentrate D = dilute Dilutions of Solutions  Dilution Equation:

Concentrated H₃PO₄ is 14.8 M. What volume of concentrate is required to make 25.00 L of 0.500 M H₃PO₄?

V_C = 0.845 L = 845 mL

Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration.

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Be sure to wear your safety glasses!

1. Measure out 0.845 L of concentrated H₃PO₄. 2. In separate container, obtain ~20 L of cold H₂O. 3. In fume hood, slowly pour [H₃PO₄] into cold H₂O. 4. Add enough H₂O until 25.00 L of solution is obtained.

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Yes; we’re OK.

You have 75 mL of conc. HF (28.9 M); you need 15.0 L of 0.100 M HF. Do you have enough to do the experiment?

>

2.1675 mol HAVE 1.50 mol NEED

M

_C

V

_C

= M

_D

V

_D

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2

1

V

M

V

M



Dilution

• Preparation of a desired solution by

adding water to a concentrate.

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Dilution

• What volume of 15.8M HNO

₃

is required to

make 250 mL of a 6.0M solution?

GIVEN:

M

₁

= 15.8M

V

₁

= ?

M

₂

= 6.0M

V

₂

= 250 mL

WORK:

M

₁

V

₁

= M

₂

V

₂

(15.8M)

V

₁

= (6.0M)(250mL)

V

₁

= 95 mL of 15.8M HNO

₃

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Preparing Solutions

How to prepare 500 mL

of 1.54 M NaCl solution

– mass 45.0 g of NaCl

– add water until total volume is

500 mL _{500 mL} volumetric flask 500 mL mark 45.0 g NaCl solute

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500 mL volumetric flask

Preparing Solutions

500 mL of 1.54M NaCl

500 mL water 45.0 g NaCl – mass 45.0 g of NaCl

– add water until total volume is 500 mL – mass 45.0 g of NaCl – add 0.500 kg of water