STRONG ACID VERSUS STRONG BASE

Acid-base titration that involves strong acid versus strong base, both the titrant and the analyte are completely ionized. The examples acid-base titrations of this type are titration between hydrochloric acid with sodium hydroxide or sulfuric acid with potassium hydroxide. The reaction and the titration curve between 0.1 M NaOH as titrant and 0.1 M HCl as analyte are describe bellow :

The H+ and OH- combine to form H2O and the other ions Na+ and Cl- remain unchanged so the net result of neutralization to a neutral solution of NaCl. The titration curve is constructed by plotting the pH of the solution as the function of the titrant added.

• In the beginning of titration there is only 0.1 M HCl so the initial pH is 1.0. to count the concentration of the H+ we use [H+] = [HCl initial]
  
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• When the 0.1 M NaOH began to add somepart of the H+ is reacted with OH- to yield H2O. So the concentration of H+ gradually decreases and the pH would be raised. The H+ concentration is calculated by [H+] = [HCl remaining]
  
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• When the equivalence point is approached (the point at which a stoichiometric reaction is complete), a neutral solution of NaCl with pH would be 7 is produced.
  
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• As we continue the the concentration of OH- rapidly increases , we then have a solution of NaOH plus NaCl. The OH- concentration is calculated with [OH-] = [excess titrant]
  

DILUTION

We often must prepare dilute solutions from more concentrated stock solutions. The millimole of stock solutions taken for dilution will be identical to the millimoles in the final diluted solution.

Example 1

You have a stock solution 0.200 M of KMnO4 and a series of 100 mL of volumetric flask. What volumes of the stock solution will you have to pipet into the flask to prepare standard solution of 0.005 M KMnO4?

Solution

Moles of 0.005 M KMnO4
= 0.005 x 100
= 0.5 mmol

We must pipet this amount from the stock solution
0.5 = 0.200 X x
    X= 0.5 / 0.200
       = 2.5 mL

Example 2

How much water should be added in 6.0 M 40 mL of H2SO4 solution to produce a 5.0 M H2SO4 solution ?

Solution

6.0 x 40 mL = 5.0 x Volume
Volume = 48 mL

The amount of water that should be added = 48 – 40 = 8 mL

NORMALITY

Useful unit of concentration in quantitative analysis is normality (N), a one normal solution contain one equivalent per liter.

An equivalent represents the mass of material providing avogadro’s number of reacting units. A reacting unit is a proton (in acid-base reaction) or an electron (in oxidation-reduction reaction). The normality of solution is calculated from:

The advantage of expressing concentration in normality and quantities as equivalent is that one equivalent of substance A will Always react with one equivalent of substance B. For the example one equivalent of KOH(=1 mol) will react with one equivalent of HNO3 (=1mol) or with one equivalent of H2SO4 (1/2 mol)

It useful to recognize that, since :

meq A = meq B

Once can calculate the volume of two solutions that will react by :

Na x mLA = NB x mLB

Example 1

Calculate the normality of the solution containing 14.205 g/L of Na2SO4 ( When SO42- reacts with two protons).

Solution

SO42- reacts with 2H+ to form H2SO4

Mol Na2SO4 per L
= 14.205 / 142.05
= 0.1 mol

Number of equivalents Na2SO4
= 0.1 x 2
= 0.2 eq

Normality = 0.2 eq / 1 L = 0.2 N

MOLARITY

Molarity of a solution is expressed as mole per liter or as millimoles per milliliter. Molar is abbreviated as M and we talk molarity of a solution when we speak of its concentration. The molarity of solution is calculated from:

Example 1

Calculate the molarity of a solution containing 10.0 g H2SO4 diluted to 250 mL?

Solution

Count the formula weight of H2SO4

= (2x1.00) + (1x32.07) + (4x16.00)
= 98.07 g/mol

The moles of H2SO4
= 10.0 / 98.07
= 0.10 mol
= 100 mmol

Molarity of H2SO4
= 100 mmol / 250 mL
= 0.4 mmol/ml = 0.4 M

MOLE

The chemist defined the mole as Avogadro’s number (6.022 x 10²³) of atoms, molecules, ions, or other species. The number of moles is calculated from:

             

Example 1

How many moles are present in 1.00 g of P4O6

Solution

The formula weight of P4O6
= (4x30.97) + (6x16.00)
= 219.88 g/mol

Mole of P4O6
= 1 / 219.88
 = 4.55 x 10^-3 mol


Example 2

Calculate the mass of 0.050 mol of dimethylnitrosamine (CH3)2N2O ?

Solution

We calculate the formula weight of dimethylnitrosamine, as follow:

= (2x12.01) + (6x1.00) + (2x14.01) + (1x16.00)
= 74.04 g/mol

Mass of (CH3)2N2O

= mol x formula weight
= 0.050 mol x 74.04 g/mol
= 3,702 g