Chapter 2, Basic Concepts

2-1 Système Internationale or SI

A system of units and prefixes established to simplify and to unify communication between scientists.

 Seven basic quantities Unit Abbreviation Mass kilogram kg Length meter m Time second s Temperature kelvin K Amount of substance mole mol Electric current ampère A Luminous intensity candela cd

Definitions of the Seven Fundamental Units:

Mass. The kilogram (kg). The kilogram of a particular cylinder of platinum-iridium alloy, called the International Prototype Kilogram, which is preserved in a vault at Sèvres, France, by the International Bureau of Weights and Measures.

Length. The meter (m). The meter is the distance that light travels in a vacuum in 1/299 792 458 of a second.

Temperature. The Kelvin (K). The Kelvin, the unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. The decision was made at the 13th General Conference on Weights and Measures on October 13, 1967 that the name of the unit of thermodynamic temperature would be changed from degree Kelvin (symbol: oK) to kelvin (symbol: K). The name (kelvin) and symbol (K) are to be used for expressing temperature intervals. The former convention which expressed a temperature interval in degrees Kelvin or, abbreviated, deg. K is dropped. However, the old designations are acceptable temporarily as alternatives to the new ones. One may also express temperature intervals in degrees Celsius.

Amount of substance. The mole (mol). The mole is the number of atoms of C12 in 12.0000 g of C12.

Electrical current. The ampère (A). The ampère (unit of electric current) is the constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular sections, and placed 1 meter apart in a vacuum, will produce between these conductors a force equal to 2 x 10-7 newton per meter of length.

Luminous intensity. The candela (cd). The candela is the luminous intensity, in the direction of the normal, of a black body surface 1/600,000 square meter in area, at the temperature of solidification of platinum under a pressure of 101,325 newtons per square meter.

Prefixes vary from yotta (factor of 1024) to yocto (factor of 10-24)

Exponent Prefixes
 Factor Prefix Symbol 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-12 pico p 10-15 femto f 10-18 atto a 10-21 zepto z 10-24 yocto y

These are the seven fundamental base units and accompanying prefixes. All of the other units may be derived from these -- they are called derived units.

Thus, 1mL = 1 cm3 and 1 Newton = 1 (kg m)/s2

2-2 The mole

A mole of molecules is Avogadro's number of molecules. Avogadro's number is the number of

12C atoms in 12.0000 g of 12C and has been experimentally determined to equal 6.022 x 10 23.

The molar mass is the mass of one mol of a substance.

Exercise 2-1a. Determine the molar mass of acetic acid, CH3COOH, .

Add up the atomic weights multiplied by the number of atoms of each element in the compound to get 60.05 g/mol.

2-3 Calculations in grams and moles

Exercise 2-2a. How many moles of citric acid, C6H8O7, MW=192.14, are contained in 6.00 g of the pure acid?

(To be performed in class)

Exercise 2-2b. How many millimoles of citric acid are contained in 6.00 g of the pure acid?

(To be performed in class)

Exercise 2-3a. How many grams Na+ are contained in 32.7 g trisodium phosphate, Na3PO4 10 H2O, F.W. 344.09?

(To be performed in class)

Exercise 2-3b. How many grams of the phosphate ion, PO43-, are contained in the amount given above?

(To be performed in class)

2-3. Solutions and their concentrations

The following concentrations and concentration relationships are of importance to and will often be found in studies involving the quantitative analyses of chemical substances:

a. Molar concentration.

b. Analytical molarity.

c. Equilibrium molarity of a particular species.

d. Percent concentration.

e. Parts per million/billion (ppm, ppb)

f. Volume ratios for dilution procedures.

g. p-functions.

Molar concentration

Exercise 2-4. Given a 1.25 liter solution which contains 3.74 g urea, H2NCONH2, MW=60.06, determine the molar concentration of urea in this solution.

(To be performed in class)

Regarding solution concentration, one speaks of analytical molarity, or that which would be present if the substance does not enter into reactions which change the concentration of the species under discussion, and equilibrium molarity of a species which reflects the disappearance of part of the original species due to dissociation, complex formation or other equilibria processes.

Exercise 2-5a. Determine the analytical molarity of 11.27 g anhydrous sodium sulfate, Na2SO4, in 500.00 mL solution.

(To be performed in class)

Exercise 2-5b. Determine the equilibrium molarity of Na+ and SO42-, written [Na+] and [SO42-], in the solution described above.

(To be performed in class)

Exercise 2-5c. A solution of acetic acid, CH3COOH, MW=60.05, having an analytical molarity of 0.100 M has what equilibrium concentration of hydronium ion, H3O+, if the dissociation goes according to the equation and the molecular form of the acid is 5% ionized?

CH3COOH + H2O <===> H3O+ + CH3COO-

(To be performed in class)

Exercise 2-5d. Determine the species molarity of CH3COO- for the solution described above.

(To be performed in class)

The student of Quantitative Analysis is often asked to prepare a solution of some given analytical molarity.

Exercise 2-6a. Explain how one would prepare 50 mL of a 0.250 M solution of sodium sulfate decahydrate,

Na2SO4 10 H2O, MW=322.19.

(To be performed in class. First the quantity is determined, then one is asked what exactly is the procedure which one follows.)

One might be asked the question above in a form which requires the weight of reagent necessary to produce a given species molarity:

Exercise 2-6b. Explain how one would prepare 50 mL of a 0.125 M solution of sodium ion, starting with the reagent sodium sulfate decahydrate, Na2SO4 10 H2O, MW=322.19, assuming 100% dissociation of sodium ion.

Percent concentration. Percent concentration may be thought of as parts per hundred.

There are three forms of percent concentration which may be encountered:

(1) wt. % (w/w) = (mass solute) ÷ (mass solution) × 100%

(2) volume % (v/v) = (volume solute) ÷ (volume solution) × 100%

(3) wt/volume % (w/v) = (mass solute) ÷ (volume solution, mL) × 100%

Weight percent is often used to express the concentration of commercial reagent grade acids. Concentrated aqueous ammonia is sold as 28% (w/w) NH3. (See the Web link

http://www.csudh.edu/oliver/chemdata/acid-str.htm

for a complete listing of properties of commercial acids and bases.)

Volume % often is used where one liquid is diluted by another:

10% aqueous butanol, for example, would be a solution in which 10 mL pure butanol is diluted to give 100 mL solution.

Wt/vol solutions are often dilute aqueous solutions in which the calculation involved takes advantage of the fact that the density of water is close to 1 g/mL. Dilute w/v solutions have concentrations very close to the values which would be reported for their w/w concentrations. At higher concentrations there is a greater divergence. For example, 50% w/w NaOH is 76.3% (w/v) NaOH because the density of the solution rises significantly above 1.00 g/mL

The use of (w/w)% for (weight/weight)% is an historical artifact which is still seen from time to time. A better designation would be (m/m)% for (mass/mass)% but because the abbreviation "m" is that which is used for the metric unit of length, (w/w) has stuck around far longer than is justifiable. To get around the use of (w/w), companies which produce reagent grade acids simply say "Assay". That is, on a bottle of Spectrum® reagent grade sulfuric acid one finds the designation "Assay (H2SO4 ). . . . . . . 95.0 - 98.0 %". That designation very definitely describes a percent concentration found by the operation

The designation (w/v) for "weight to volume" ought also to be "(m/v)" for a description of the operation

Exercise 2-7. Knowing the above values for the (w/w) and (w/v) solution of NaOH, determine the density of that solution.

(To be performed in class)

Consider the following table of solution densities at 20oC:

 Solution Density 10% (w/w) KCl 1.08 g/cc 10% (w/w) NaCl 1.07 g/cc 10% (w/w) KOH 1.08 g/cc 10% (w/w) KCl 1.06 g/cc 10% (w/w) BaCl2 2H2O 1.082 g/cc

Exercise 2-8. Determine the (w/v) value for the 10% (w/w) solution of KOH above.

(To be performed in class)

Parts per million (ppm). Parts per million is a concentration convention convenient to use for very dilute solutions.

Exercise 2-9. Determine the ppm of ferrous ion, Fe2+, in a solution known to be 1.2 × 10-6 M Fe3(PO4)2 8H2O

FW=501.61.

(To be performed in class)

Volume Ratios for Solutions to be Diluted

When diluting solutions, a directive may specify a 1:5 volume/volume dilution. Although this usually means 1 volume of concentrated solution to 5 volumes of the final dilute solution, that might not be the case depending upon the procedure in any particular laboratory. The student is advised always to verify whether the meaning might be 1 volume of concentrated solution to 5 volumes of water. Making a mistake in the preparation of the final solution can lead to serious consequences particularly where the preparation of nutrient solutions in experimentation on humans and animals is involved.

Exercise 2-10. Determine the volume/volume ratio needed to dilute a 0.5 M saline solution to 0.2 M.

(To be performed in class)

p-functions The p-function is the negative logarithm to the base 10 of the molarity of a given species.

Thus, pX = -log[X]

Exercise 2-11a. Given a solution known to be 3.7 × 10-4 M NaCl and 4.5 × 10-3 M HCl , determine pH, pNa and pCl for these solutions.

(To be performed in class)

Or you may be asked to do the reverse.

Exercise 2-11b. Given a solution with a pOH of 4.37, determine the species molarity of the OH- ion.

(To be performed in class.)

Notes on logarithmic calcuations:

Today with modern hand calculators, converting X to log10X is a matter of a single keystroke. Converting a p function to a concentration is equally simple because of the following relationship.

Exercise 2-12. Consider pAg in a solution of silver nitrate, AgNO3, to equal 6.74. What is the molarity of Ag+ ?

6.74 = pAg = -log10[Ag+ ]

Since the statement y=log10X could be identically rewritten as X=10y, the expression above could also be rewritten as

[Ag+ ] = 10-6.74

To convert to molarity, enter -6.74, and press the 10X key, to get

[Ag+ ] = 10-6.74 = 1.82 × 10-7

Density and specific gravity of solutions.

The numeric value for a density given as g/mL or kg/L is the same. Water with a density at 20oC of 0.998 g/mL has a density of 0.998 kg/L. The density of gases is usually given as g/L because at normal pressures the density is about a thousand times less than that of liquids. The density of dry air at 760 torr at 1 atm pressure is 1.185 g/L.

The specific gravity is a ratio between the mass of a given volume and the mass of an equal volume of water at 4oC. Since the density of water at 4oC is practically 1.00 g/cc, the density and specific gravity of aqueous solutions are almost identical. The difference needs always to be considered for analytical work of high precision, however.

Exercise 2-13. Commercial reagent grade formic acid is 90% w/w and has a specific gravity of 1.20. Determine the molarity of formic acid in a bottle of commercial reagent grade formic acid.

(To be performed in class)

Given any two of the following three: (w/w)%, molarity and specific gravity, one can calculate the third. Given just the molarity, one is able to calculate methods of diluting solutions volumetrically.

Exercise 2-14. Hydriodic acid, HI, is sold in reagent grade bottles as 57% (w/w) HI and has a specific gravity of 1.70. Explain how you would prepare 100 mL of 1.25 M hydriodic acid, using only a 100 mL volumetric flask and a 20 mL graduated pipette.

(To be performed in class)

Stoichiometric calculations.

Stoichiometric calculations follow part or all of the path

1. Mass x ---> 2. Moles x ---> 3. Moles y ---> 4. Mass y

Going from step 2. to step 3. requires a stoichiometric ratio which is usually determined by balancing a chemical equation.

Exercise 2-15a. Calcium oxalate, CaC2O4 , can be precipitated by adding to a solution of calcium hydroxide, Ca(OH)2, sufficient sodium oxalate, Na2C2O4 , to react with all calcium hydroxide present. What mass sodium oxalate would be required to precipitate all of the calcium in a solution containing 3.74 g Ca(OH)2?

(To be performed in class)

Exercise 2-15b. What weight of Calcium oxalate, CaC2O4, would be produced?

(To be performed in class)

Exercise 2-16a. Limiting reagent.

25.0 mL 0.100 M Ca(OH)2 is added to 20.0 mL 15.8 % (w/v) Na2C2O4. Calculate the weight CaC2O4 formed.

(To be performed in class)

Exercise 2-16b. What is the equilibrium concentration in molarity of all ionic species remaining after the reaction?

(To be performed in class)