in water and produce proton of hydrogen H+.
Salts
When an acid reacts with a base, a salt is produced.
NaOH + HNO3 = NaNO3 + H2O
To calculate the percentage composition of NaNO3, find the molecular mass of NaNO3
23 (Na) +12 (N) +16*3 (O3) = 83
The molecular weight of NaNO3 = 83
83 – 100%
23 (Na) – X%
X = 23 * 100 /83 = 27.7% of Na
83 – 100%
12 (N) – X %
X = 12 * 100/83 = 14.5% of N
83 – 100%
48 (O) – X %
X = 48 * 100/83 = 57.8% of O
Some salts are more soluble in water; some are less soluble or not soluble at all. When a non-soluble salt is produced as a result of an acid and base reaction, a precipitate is formed.
Ca (OH) 2 + H2CO3 = CaCO3 +2H2O
CaCO3 is not soluble in water. A white precipitate is formed.
Salts may react with each other and new salts are produced:
Equivalent proportions
All chemical reactions occur in equivalent proportions.
1. How many grams of Ca Cl2 are spent in the following reaction:
10g?g
Na2CO3 + Ca Cl2 = CaCO3 +2NaCl
All compounds react with each other in certain proportions. In a given reaction one mole of Na2CO3 reacts with one mole of CaCl2
A mole is MW (Molecular mass) in grams.
For Na2CO3 MW is 23 *2 +12 +48 = 106 g. = 1 mole
For CaCl2 MW is 40 +35*2 = 110g. = 1 mole.
106 g Na2CO3 react with 110g CaCl2
10g Na2CO3 react with X g CaCl2
X= 10 * 110 / 106 = 10.38 g CaCl2
2. How many grams of CaCO3 are produced if 100 ml of 0.5 M solution of Na2CO3 reacts with an unlimited volume of solution CaCl2?
0.5 M
2. Na2CO3 + Ca Cl2 = CaCO3 +2 NaCL
100 ml
1 liter of I M solution of Na2CO3 contains 106 g
How many grams of Na2CO3 in 1 liter of 0.5 M solution?
1 M – 106 g
0.5 M – X g
X = 0.5 M * 106 g / 1 M = 53 g
I liter of 0.5 M solution contains 53 grams of Na2CO3
How many grams of Na2CO3 are in 100 ml?
1 liter – 53 g
0.1 liters – X g
X = 0.1 * 53 /1 = 5.3 g
106 g of Na2CO3 produces 100 g of CaCO3
5.3 g of Na2CO3 produces X g of CaCO3
X= 5.3 * 100 / 106 = 5 g
Acid Base reactions
Molarity. Molality. Normality
1. Let say we have 100 ml of H2SO4 solution and it contains 0.49 g of H2SO4
What is Molarity?
What is Molality?
What is Normality?
1 Molar solution contains a number of grams equal to molecular mass per one liter of solution.
MW of H2SO4 = 2 +32 +4* 16 = 98g
I mole = 98g/L
We have to find how many grams of H2SO4 given solution are contained in 1 liter.
100 ml = 0.1 L and contains 0.49 g
1 L contains X g
X = 0.49 * 1 / 0.1 = 4.9 g.
98g/L – 1 mole
4.9 g/L – X moles
X = 4.9 * 1 / 98 = 0.05 moles
The molarity of a 100 ml solution of H2SO4, which contains 0.49 g of H2SO4, equals 0.05 moles.
What is Molality? Molality is moles of solute / kg of solvent.
A solute is a substance dissolved in another substance.
A solvent is a substance in which another substance is dissolved
What is normality? An equivalent is the molecular mass or mass of acid or base that produce one mole of protons (H+) or one mole of hydroxyl
(OH-) ions.
One mole of H2SO4 produces 2 moles of H+ then equivalent to H2SO4 = MW/2 = 49g/L
49g/L is 1 Normal solution
4.9g/L solution is – X N
X= 4.9 * 1/ 49 = 0.10 N.
The normality of 100 ml solution of H2SO4, which contains 0.49 g of H2SO4, equals 0.10 N.
2. We have 10 ml of NaOH unknown concentration. The solution was titrated with 0.10 N solution of H2SO4 and 15 ml were required for neutralization. What is the concentration of NaOH?
Nb * Vb = Na * Va
where V is volume, N is Normality, b – base and a – acid
Nb = Na * Va / Vb = 0.10 * 15 / 10 = 0.15 N
The concentration of NaOH = 0.15 N.
Weight and Volume problems:
1. How many liters of Hydrogen are produced from one liter of water?
2H20 = 2H2 +02
First, we have to find how many grams of water are spent and how many grams of H2 are produced?
MW
