INSMETHLecture 1: IntroductionLecture 1: Introduction

Ma’am Glenn Medina

De La Salle University

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CONCENTRATION

UNITS: REVIEWUNITS: REVIEW

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SIGNIFICANT FIGURES

�Significant figures are meaningful or important digitsin a measured quantity.

�RULES:

1. All nonzero digits are significant.

2. Zeros between nonzero digits are significant

3. For nos.<1, all zeros to the right of the first3. For nos.<1, all zeros to the right of the firstnonzero digit are significant. All zeros to the left of thefirst nonzero digit are not significant.

4. For nos.>1, all zeros to the right of the decimalpoint are significant.

5. For large numbers that do not contain digits afterthe decimal point, the terminal zeros may or may notbe significant.

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Exercise – Significant Figures

1. 25 6. 0.00005642

2. 0.09034 7. 3000

3. 1.00 8. 2.056

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3. 1.00 8. 2.056

4. 0.50789 9. 3.654 x 104

5. 2008 10. 4.0570

SIG FIG – OPERATIONS

ADDITION AND SUBTRACTION

�The limiting or key number is the

measurement with the least number of

decimal places (or least number of digits

specified after the decimal point). This indicates

the measurement obtained with the leastthe measurement obtained with the least

degree of precision.

�After performing the indicated mathematical

operation, the final answer must be rounded

off to contain the same number of digits afterthe decimal point as the limiting number.

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ADDITION AND SUBTRACTION – Examples

423.1

3692.3

+ 7793.8

6.95 tooff round

9463.6

20.2+

6.671 tooff round

6707.6

312.2−

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MULTIPLICATION AND DIVISION�The limiting number is the measurement with the

least number of significant figures.

�After performing the indicated mathematical operation,

the final answer must be rounded off to contain the

same number of significant figures as the limitingsame number of significant figures as the limiting

number.

5.22 tooff round

21766.5

31.2x

224.4

3.9 tooff round

89648.3

41.x

2783.2

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EXACT NUMBERS

Exact numbers have infinite (very large number)

number of significant figures. These numbers are

usually not considered in determining the limiting

number in an operation. Exact numbers areobtained from

Counting numbers 10 marbles 5 dozens of eggs

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Counting numbers 10 marbles 5 dozens of eggs

Number of data or

experimental measurements

Average = (3.25 + 3.20 + 3.22)

3Mean = (0.1044 + 0.1058)

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Result of conversions

1.025 g/mL is written as 1.025 g = 1 mL

44.01 g = 1 mole

Definition of calculated values

Percent –ratio is multiplied by 100

Parts per million –

involves the factor of 106

LOGARITHMS AND ANTI-LOGARITHMS

The number of significant figures in the mantissa of the

original number must be the same as that in the

number in the logarithm.

Ex: log (9.57 x 104) = 4.891

For antilogarithms, the total number of SF in the

original number must be the same as the number

of SF in the mantissa of the number in the

antilogarithm.

Ex: antilog 4.891 = 9.57 x 104

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Concentration Units

I. MolarityII. MolalityIII. Percent concentrationIV. ppm, ppbV. pH, pXV. pH, pXVI. DensityVII. Specific gravityVIII.Titer

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I. molarity

• MOLARITY (M) = moles of solute per Liter of solution(mol/L) OR millimoles of solute per milliliter ofsolution (mmol/mL)

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Since moles = mass / molar mass

I. MOLARITY

Since moles = mass / molar mass

Then,

Mass = Molarity * Molar Mass * Vol (L)12

1. M = n/V

2. m = M*MM*V

I. MOLARITY - EQUATIONS

2. m = M*MM*V

3. (MV)conc = (MV)dil

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1. Calculate the molar concentration of a 250.00 mL solution containing 0.0524 mol NaCl.

2. How many moles of NaCl are dissolved in 15.23 mL of 0.124 M NaCl solution?

3. How many mL of 0.120 M must be measured

I. MOLARITY – SAMPLE PROBLEMS

3. How many mL of 0.120 M must be measured in order to obtain 0.2548 g of NaCl (M.M. = 58.44)?

4. How many grams of NaCl must be dissolved in order to prepare 500.0 mL of 0.120 M NaCl?

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answers

1. M = 0.0524 mol / 0.25000 L = 0.210 M

2. Moles = 0.124 M*0.01523 L = 0.00189 mol

3. Vol (mL) = 0.2548 g / (58.44 g/mol *0.120 M) =3. Vol (mL) = 0.2548 g / (58.44 g/mol *0.120 M) =

0.0363 L or 36.3 mL

4. Mass = 0.120 M* 58.44 g/mol *0.5000 L = 3.51 g

Preparing solutions

1. How many grams of Copper (II) sulfate

pentahydrate should be dissolved in a

volume of 500.0 mL to make 8.00 mM

Cu+2? MM = 249.69 g/nCu+2? MM = 249.69 g/n

2. How many mL of conc HCl must be diluted

to 1.000 L to make 0.100 M HCl solution?

Preparing solutions

1. How many grams of Copper (II) sulfatepentahydrate should be dissolved in avolume of 500.0 mL to make 8.00 mMCu+2? MM = 249.69 g/n Ans: 0.999 g

2. How many mL of conc HCl must be dilutedto 1.000 L to make 0.100 M HCl solution?Ans: 8.26 mL

ii. MOLALITY

• Moles of solute per kilogram of SOLVENT

• Independent of temperature• Independent of temperature

• Used in colligative properties

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iii. PERCENT CONCENTRATION

1. Weight percent = (wt solute/wt soln)*100

2. Volume percent = (vol solute/vol soln)*100

3. Wt/Vol percent = (wt solute/vol soln)*1003. Wt/Vol percent = (wt solute/vol soln)*100

• Ex: A rubbing alcohol with 70%(v/v) IPA

contains 70 mL IPA in 100 mL solution.

iii. % CONC – sample problems

1. Calculate the mass of solute present in a

35.0%(w/w) solution containing 250.0 g of

solvent.

2. A sample of impure NaCl was found to contain

25.00% NaCl. How many grams of NaCl is25.00% NaCl. How many grams of NaCl is

present for every 100.00 g of the sample?

3. A sample of impure PbS was found to contain

20.0% PbS. If 3.250 g of this sample was

obtained, how many grams of PbS are present in

the sample?

answers

1. Calculate the mass of solute present in a

35.0%(w/w) solution containing 250.0 g

of solvent. 35.0 = (m/(m+250.0))*100

m = 135 gm = 135 g

2. 25.00 g NaCl

3. 0.65 g PbS

CONVERTING WT % TO MOLARITY/MOLALITY

1. Find the molarity and molality of 37.0

wt% HCl. The density is 1.19 g/mL. MM

of HCl = 36.46

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answers

12. 1 M HCl

16.1 m HCl16.1 m HCl

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iv. Ppm and ppb

• Parts per million (ppm) =

mg/kg or mg/L

• Parts per billion (ppb) =

µg/kg or µg/L

iv. Ppm and ppb – sample problems

1. A solution of Cu(NO3)2 contains 55.8

ppm of Cu+2. Convert this to ppb. What

is the molarity of Cu+2 in this solution?

2. What is the ppm of K+ in a K3Fe(CN)62. What is the ppm of K+ in a K3Fe(CN)6

solution with a molarity of 5.77x10-4?

3. Calculate the ppm of Ca in a 25.00 mL

water sample that contains 82.4 mg of

Ca.

answers

1. A solution of Cu(NO3)2 (MM = 155.57 g/mol)

contains 55.8 ppm of Cu+2. Convert this to

ppb. What is the molarity of Cu+2 in this

solution?

55800 ppb Cu+2ppb = 55.8 ppm*1000 = 55800 ppb Cu+2

M = 55.8 mg/L * 1g/1000mg * 1mol/63.55 g = 8.78 x

10-4 M Cu+2

2. 67.7 ppm

Convering ppb into molarity

1. The concentration of C29H60 in summer

rainwater collected in Germany is 34

ppb. Find its molarity.

Answer = 8.3 x 10-8 M or 83 nM

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V. Ph and px

1. pH = -log [H+]

2. pX = -log [X]

V. Ph and px – SAMPLE PROBLEMS

1. Calculate the molarity of Ag+ in a

solution with a pAg of 6.372.

2. What is the pH of a solution

having [H+] = 1.0x10-7?having [H+] = 1.0x10-7?

3. Calculate the pNa+, pCl- and pOH-

in a solution that is 0.0335 M in

NaCl and 0.0503 M in NaOH.

Ph and px - exercise

1. Calculate the molarity of

Ag+ in a solution with a pAg

of 6.372. Ans: 4.25x10-7of 6.372. Ans: 4.25x10

2. What is the pH of a solution

having [H+] = 1.0x10-7? Ans:

7.00

VI. Density and specific gravity

o DENSITY = g/mL

o SPEC GRAVITY =o SPEC GRAVITY =

Dsubstance / DH2O

(at same Temp)

VI. Density and SG – SAMPLE PROBLEMS

1. What is the density of a

substance Y if its mass is 25.07 g

and it occupies a volume of

17.88 cm3.17.88 cm3.

2. Give the specific gravity of the

substance Y if the density of H20

is 0.9973 g/mL.

ANSWERS

1. D = 25.07 g/17.88 mL = 1.402

g/mL

2. SG = 1.402 g/mL /0.9973 g/mL =

1.406

VII. NORMALITY

• The Normality of a solution is the

number of equivalents of solute

dissolved per liter of solution.

• Unit = equivalents solute / liter • Unit = equivalents solute / liter

solution

• Example: 1.0 N is read as 1.0

normal

VII. NORMALITY

• N = h*M where h (eq/mol)

depends upon the nature of the

substance involved

• For acids, h = # of H+ given off• For acids, h = # of H+ given off

• For bases, h = # of OH- given off

• For substances in redox reactions,

h = # of electrons in the redox

reaction

VII. NORMALITY – SAMPLE PROBLEMS

1. What is the normality of a

solution that is 0.454 M in

H2SO4?H2SO4?

2. Give the molarity of 0.0879

N MnO4- solution (5 e-s).

ANSWERS

1. N = 2 eq/mol* 0.454 mol/L

= 0.908 N or eq/L

2. M = N/h = 0.0879 eq/L / 5 2. M = N/h = 0.0879 eq/L / 5

eq/mol = 0.0176 eq/L

STOICHIOMETRY CALCULATIONS

• Iron from a dietary supplement tablet can be

measured by dissolving it and then converting

the Fe into Fe2O3.

• Gravimetric Analysis steps:

– Tablet (FeC4H2O4) mixed with HCl. Filter.

– Oxidize Fe+2 into Fe+3 with H2O2

– NH4OH added. Heat in a furnace to form Fe2O3

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HOMEWORK

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1-281-28

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1-35

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