Basic Tools of Analytical Chemistry (alat dasar untuk kimia analitik)

Basic Tools ofAnalytical Chemistry

By: Isman Kurniawan

Thursday, February 6, 14

Fundamental units of measure

• mass

• volume

• distance

• time

• temperature

• current

• amount of substance


Scientific notation

• 1 mole contain 602.213.670.000.000.000.000.000 particles

• 1 mole = 6,02 x 1023


Scientific notation


CONCENTRATION

• the relative amount of solute per unit volume or unit mass of solution

concentration =

amount of solute

amount of solution


CONCENTRATION

• Molarity (M)

• the number of moles of solute per liter of solution

FW = formula weight

M =

moles solute

liters solution

=

grams solute

FW

⇥ 1

liters solution


CONCENTRATION

• Normality (N)

• the number of equivalents of solute per liter of solution

• the number of equivalents (n): part of chemical species involved in a reaction

• example:

a. H2SO4(aq) + NH3(aq) HSO4-(aq)+ NH4+(aq) (n=?)

b. H2SO4(aq) + 2OH-(aq) SO42-(aq)+ 2H2O(l) (n=?)


CONCENTRATION

• Normality

EW =FW

nN = M ⇥ n

EW = equivalent weight

N =

moles equivalent solute

liters solution

=

grams solute

EW

⇥ 1

liters solution


CONCENTRATION

• Normality

• example-1:Calculate the equivalent weight and normality for a solution of 6.0 M H3PO4 givent the following reactions:

a. H3PO4(aq) + 3OH-(aq) PO4-(aq) + 3H2O(l)

b. H3PO4(aq) + 2NH3(aq) HPO42-(aq) + 2NH4+(aq)

c. H3PO4(aq) + F-(aq) H2PO4-(aq) + HF(aq)


CONCENTRATION

• Molality (m)

• the number of moles of solute per kilogram of solvent

m =

moles solute

kg solvent

• temperature independent unit of concentration


CONCENTRATION

• Weight percent (%w/w)

• grams of solute per 100 g of solution

%w/w =

g solute

100 g solution


CONCENTRATION

• Volume percent (%v/v)

• milliliters of solute per 100 mL of solution

%v/v =

mL solute

100 mL solution


CONCENTRATION

• Weight-to-volume percent (%w/v)

• grams of solute per 100 mL of solution

%w/v =

g solute

100 mL solution


CONCENTRATION

• Part per million (ppm)

• milligrams of solute per liter of solution

ppm =

mg solute

liters solution

=

g solute

10

6mL solution


CONCENTRATION

• Part per billion (ppb)

• micrograms of solute per liter of solution

ppb =

µg solute

liters solution

=

g solute

10

9mL solution


CONCENTRATION

• Converting between concentration units

• example-2:A concentrated solution of aqueous ammonia is 28.0% w/w and has a density of 0.899 g/mL. What is the molar concentration of NH3 in this solution?[14,8 M]


CONCENTRATION

• Converting between concentration units

• example-3:The maximum allowed concentration of chloride in a municipal drinking water supply is 2.50 x 102 ppm Cl-. When the supply of water exceeds this limit, it often has a distinctive salty taste. What is the concentration in moles Cl-/ liter?[7.05 x 10-3 M]


CONCENTRATION

• p-Functions

• a function of the form pX

pX = � log(X)


CONCENTRATION

• p-Functions

• convenient to express small changes of concentration


CONCENTRATION

• p-Function

• example-4What is pNa for a solution of 1.76 x 10-3 M Na3PO4?

• example-5What is the [H+] in a solution that has a pH of 5.16?


STOICHIOMETRIC

• example-6The amount of oxalic acid in a sample of rhubarb was determined by reacting with Fe3+ as following reaction:2Fe3+(aq) + H2C2O4(aq) + 2H2O(l) 2Fe2+(aq) + 2CO2(g) + 2H3O+(aq)

In a typical analysis, the oxalic acid in 10.62 g of rhubarb was extracted with a suitable solvent. The complete oxidation of the oxalic acid to CO2 required 36.44 mL of 0.0130 M Fe3+. What is the weight percent of oxalic acid in the sample of rhubarb?[0.201%]


STOICHIOMETRIC

• example-7One quantitative analytical method for tetraethylthiuram disulfide, C10H20N2S4 (antabuse), requires oxidizing the sulfur to SO2, and bubbling the resulting SO2 through H2O2 to produce H2SO4. The H2SO4 is then reacted with NaOH according to the reaction:H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)What is the weight percent C10H20N2S4 in a sample of antabuse if the H2SO4 produced from a 0.4613 g portion reacts with 34.85 mL of 0.02500 M NaOH?[7.001%]


PREPARING SOLUTIONS

• Stock solutions

• a solution of known concentration from which other solutions are prepared

• example-8:Describe how you would prepare the following three solutions: (a) 500 mL of approximately 0.20 M NaOH using solid NaOH; (b) 2 L of 4% v/v acetic acid using concentrated glacial acetic acid.


PREPARING SOLUTIONS

• Dilution

• the process of preparing a less concentrated solution from a more concentrated solution

no

= nd

Co

⇥ Vo

= Cd

⇥ Vd


PREPARING SOLUTIONS

• Dilution

• example-9:A laboratory procedure calls for 250 mL of an approximately 0.10 M solution of NH3. Describe how you would prepare this solution using a stock solution of concentrated NH3 (14.8 M).


PREPARING SOLUTIONS

• Dilution

• example-10:A sample of an ore was analyzed for Cu2+ as follows. A 1.25 g sample of the ore was dissolved in acid and diluted to volume in a 250 mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipet to a 50 mL volumetric flask and diluted to volume. An analysis showed that the concentration of Cu2+ in the final solution was 4.62 ppm. What is the weight percent of Cu in the original one?[0.231%]


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Basic Tools of Analytical Chemistry (alat dasar untuk kimia analitik)