Comparing the Concentration of Some Alkalis in Saturated Solutions.

Chemistry Planning Exercise:

Comparing the Concentration of Some Alkalis in Saturated Solutions:

Introduction:

When metals from groups one and two of the periodic table combine with an OH molecule, the type of intramolecular bonding is ionic bonding. The metals lose one or more electrons and so become positively charged. The non-metal ions/ molecules gain electrons and so become negatively charged. The oppositely charged ions attract each other to form a rigid 3-D lattice a. Each ion in the lattice is surrounded by others of opposite charge e.g. NaOH:

Due to their polar nature, ionic compounds are usually soluble in polar solvents, for example water:

Water is polar due the difference in electronegativity between the oxygen and hydrogen atoms. This is represented by δ+ and δ-.

When an ionic substance dissolves in a solvent, the lattice must be broken and the separated species must then be surrounded by individual solvent molecules b. The solubility of a solute in water, at a given temperature is the maximum amount of it that will dissolve in 100 grams of water at that temperature. When this amount is reached, it gives a saturated solution c. In a saturated solution, e.g. Ca(OH)2 (aq), Ca2+ and OH- ions are in equilibrium with the solid calcium hydroxide:

Ca(OH)2 (aq) Ca2+ + 2OH-

The solubility of group one and two hydroxides varies greatly, however, clear trends can be seen:

(303K) = at a temperature of 303K or 30oC

Clearly, as you move down the group, the solubility increases and solubility decreases as you move across a period. This is due to the size of ions and their reactivity.

As you move down groups one and two, the atomic radii of the atoms increases. This is because:

The number of electron shells is increasing;
Distance of outer electron from nucleus increases;
Increases shielding of positive nuclear charge ∴outer electrons held less tightly;
Increased electron repulsion.

The increases size of metal ions at the base of groups one and two means that the distance between OH- and X+/2+ ions will be greater, therefore the electrostatic attraction between them is less. Subsequently the solid 3-D lattice will break apart much more easily and the hydroxide will be more soluble than those with smaller metal ions b.

As you move across a period, the solubility decreases. This is due to a decreased atomic radius and an increased charge on the ions. With a smaller atomic radii, the distance between OH- and metal ions decreases, thus the electrostatic attraction is ...