1. Formation and dissolution of a precipitate

Take 3-4 drops of a solution of any calcium salt into a test tube and add a solution of ammonium oxalate until the formation of a precipitate. Dissolve the precipitate in hydrochloric acid. Compare the values of the dissociation constants of oxalic acid and the solubility product constant for calcium oxalate and make your conclusion.

2. Direction of a chemical reaction

a) Take 3-4 drops of any barium salt in a test tube and add potassium chromate until the formation of a precipitate. Divide the precipitate into two parts and add HCl or CH₃COOH respectively. Notice your observations.

b) Take 3-4 drops of any copper salt in one test tube and 3-4 drops of any manganese salt in another test tube. Add sodium sulfide until metal sulfide precipitates is formed. Try to dissolve each precipitate in a diluted solution of hydrochloric acid.

Calculate equilibrium constants of the reactions of dissolution of precipitates and make your conclusion about the possibility of each reaction.

3. Heterogenous equilibrium

a) Mix 5-6 drops of silver nitrate with 2-3 drops of sodium chloride in a test tube. Observe the formation of a precipitate of AgCl. Then add 2-3 drops of potassium chromate into the same test tube. Stir the mixture and observe the change in color of the precipitate. Add 2-3 drops of sodium sulfide and write down the observations.

b) Repeat the experiment changing the order of addition of reactants. Add first NaCl, then Na₂S and finally K₂CrO₄s.

Calculate the equilibrium constants of all the reactions and make your conclusion about the shifting of chemical equilibria.

Questions and problems

1. Calculate the solubility product constant of PbBr₂ if the solubility of the salt is 1.3210 molel^-1.

2. Will a precipitate of silver sulfate be formed if a 1M solution of H₂SO₄ is added to an equal volume of a 0.02 M solution of AgNO₃?

3. Calculate the equilibrium constant of the reaction and explain whether the precipitate of calcium oxalate can be dissolved in acetic acid.

Experiment 3

Ionic product of water. Ph. Hydrolysis of salts

Water is a very weak electrolyte and dissociates in a very small extent, forming hydrogen ions and hydroxide ions:

H₂O H⁺ + OH

The equilibrium state of this reaction can be characterized by a constant called ionic product of water:

K_W = [H⁺][OH] = 10 (at 22^oC)

Instead of concentrations of H⁺ and OH ions, it is more convenient to use their common logarithms taken with the reverse sign; these quantities are denoted as pH and pOH:

pH = - log[H⁺]; pOH = - log [OH]

Solutions in which concentrations of hydrogen and hydroxide ions equal each other are called neutral solutions, [H⁺] = [OH] = =10, pH = 7. In acidic solutions, [H⁺] > [OH], pH < 7; in alkaline solutions [H⁺] < [OH], pH > 7.

When we dissolve a salt comprising an anion of a weak acid or a cation of a weak base in water, a hydrolysis process occurs, i.e. an exchange reaction between salt and water, the result of which is the formation of a weak acid or a weak base.

The process of hydrolysis can be quantitatively described using the notions of hydrolysis constant (K_h) and degree of hydrolysis (h).

If a salt is formed by a weak acid and a strong base, the anion of a salt undergoes hydrolysis, hydroxide ions are formed in the solution:

NaCN + H₂O NaOH + HCN

CN^- + H₂O OH^- + HCN; pH > 7

K =

In the hydrolysis of a salt formed by a strong acid and a weak base, the cation of the salt becomes hydrolyzed, the solution becomes acidic:

NH₄Cl + H₂O NH₄OH + HCl

NH₄⁺ + H₂O H⁺ + NH₄OH; pH < 7

K_h =

When a salt formed by a weak acid and a weak base reacts with water, both its cation and its anion become hydrolyzed. In this case, the reaction of the solution depends on the relative strengths of the acid and the base forming the salt:

NH₄CH₃COO + H₂O NH₄OH + CH₃COOH

NH₄⁺ + CH₃COO + H₂O NH₄OH + CH₃COOH

K_h =

The hydrolysis of salts formed by weak polybasic acids or weak polyvalent metal ions proceeds stepwise. Under common conditions, only first-step hydrolysis process should be taken into consideration:

K₂CO₃ + H₂O KHCO₃ + KOH

CO₃² + H₂O HCO + OH

AlCl₃ + H₂O AlOHCl₂ + HCl

Al³⁺ + H₂O AlOH²⁺ + H⁺

Degree of hydrolysis “h” is defined as a fraction of an electrolyte that has become hydrolyzed. It is related to the hydrolysis constant by an equation similar to the Ostwald dilution law for the dissociation of a weak electrolyte:

K_h =

Most often, the hydrolyzed part of a salt is very small, h << 1. In this case

K_h = h²C

The last equation shows that the degree of hydrolysis of a given salt increases when its concentration diminishes.

The equilibrium of a reaction of hydrolysis can also be shifted by change in temperature. As it is an endothermic process, the rise of temperature leads to the increase of hydrolysis.

If we introduce a reagent combining with the H⁺ or OH ions formed in hydrolysis into a solution of a hydrolyzing salt, in accordance with Le Chatelier’s principle equilibrium will shift towards the direction of an intensification of hydrolysis; as a result, hydrolysis may go up to the end, leading to formation of its products. The H⁺ (or OH) ions can be combined together to form water molecules by introducing another salt whose hydrolysis leads to the accumulation of OH (or H⁺) ions in the solution; the H⁺ and OH ions will neutralize one another, this causing the mutual intensification of the hydrolysis of both salts and the formation of the hydrolysis products.

EXPERIMENTAL PART