4. Solution Equilibria PDF

Preview

Summary

Chem notes...

Description

Inquiry Question: How does solubility relate to chemical equilibrium. Describe and analyse the processes involved in the dissolution of ionic compounds in water. Dissolution process of Ionic Compounds: - Soluble ionic compounds will dissolve in water to form an aqueous solution. - When a soluble ionic compound is added to water, the ions at the surface of the crystal become surrounded by water molecules. - This force is known as ‘Ion-dipoles’ - When the ion-dipole forces (adhesive forces) are greater than the ionic bonds between the ions and the hydrogen bonding within the water (cohesive forces), the ions are dislodged from their positions in their crystal. - The ionic compound dissociates into its component ions. Thus, the ions become solvated. - The solvated or hydrated ions are surrounded by a shell of water molecules known as the solvation layer. - The solvation layer acts as a cushion and prevents a solvated anion from colliding directly with a solvated cation, and therefore keeps in the ions in solution. - These ions are able to move freely throughout the aqueous solution, hence being able to conduct electricity. - For ionic compounds that has unequal numbers of ions, the ratio of the dissociated ions produced is shown using coefficients in front of their formulae. For example, the balanced chemical equation fo the dissolution of MgCl2 is written as:

−¿ ¿ 2+¿+2 Cl ( aq ) ¿ MgCl2(g ) → Mg( aq) Note: Each ionic compound dissolves to form one type of cation and one type of anion only. -

-

Not all ionic compounds are able to dissolve due to their insolubility, e.g. AgCl and Ca3( PO 4 ) 2 . The insolubility of these ionic compounds is due to their strong ionic bonds which cannot be disrupted by the adhesive ion-dipole forces between solute ions and water molecules. Hence, the ions do not become dislodged from their positions in the crystal. Entropy can also contribute to insolubility. Most dissolutions are entropically favourable, but some dissolutions are unfavourable.

Example: Outline the changes that occur in bonding and entropy when sodium bromide dissolves in water. -

When sodium bromide is dissolved in water it forms an aqueous solution of

+¿¿ −¿ ¿ Na(aq) and Br (aq) .

−¿ ¿ +¿+ Br(aq) . Due to sodium bromide being an ionic compound that is added to water, the ¿ NaBr(s) → Na(aq ) +¿ −¿ ¿ ¿ Na(aq) and Br (aq) ions become surrounded by water molecules. This force is known as ion-dipoles (adhesive force). These ion-dipole forces are greater than the hydrogen bonding within the water (cohesive force), the and

+¿ −¿ ¿ and Br¿ ions are dislodged from their positions from their crystals. The Na(aq) (aq)

+¿ ¿ Na(aq)

−¿ ¿ Br (aq) ions are now surrounded by a shell of water known as the solvation layer. The solvation layer

acts as a cushion and prevents solvated ions from colliding with solvated cations, therefore keeping the ions

+¿¿ Na(aq)

−¿

and Br¿ ions are now able to freely move through the solution thus (aq) increasing the overall entropy of the system. separate in solution. The

Investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when removing toxicity from foods, for example: - Toxins in cycad fruit

Detoxification process: - Several physical and chemical process are commonly used for detoxification. Process Effect Cooking/Roasting - The food items are heated in an oven or fire. - This causes the toxins within the food to decompose. Leaching - The food items are cut-up then soaked in water which is changed multiple times. They could also be left in running water. - Water-soluble toxins are washed away. Fermentation/Prolonged storage

-

The food items are stored for long periods (months to years) During this time various biological processes occur that break down the toxins e.g. eaten by fungi, broken down by plant’s natural enzymes (Biological catalysts).

Cycads: - Cycads are palm-like plants that produce seeds in cones. - Cycad seeds are a rich source of carbohydrates however they are highly toxic. The main two toxins found are cycasin and BMAA. - Eating without detoxification led to gastrointestinal disturbances, ataxia (loss of muscle control), liver damage, possible dementia and cancer. - The indigenous people who ate these seeds would detoxify them. - The seeds could be cooked in an oven or fire. - The seeds could be cut open or ground up, and then leached in water. The seed pieces were usually placed in mesh bags or woven baskets and submerged in running water for around a week. - A similar method included soaking the seeds multiple times, draining the water each time. - Fermentation or storage for several months, or several years. - The remaining starchy paste after soaking would then be eaten without no further cooking. It could also be dried, ground up and formed into loaves that were then baked in bread. Black bean: - The Moreton Bay Chestnut (Castanospermum australe), also known as black bean, is found on the east coast of Australia. It produces pods that contain large seeds that are toxic. - Eating unprocessed seeds caused vomiting and diarrhoea. Cooked, processed seeds tasted like sweet chestnuts. - The detoxification process includes roasting, breaking up the seeds and leaching them in water, before grinding into flour and shaping into loaves for cooking.

Conduct an investigation to determine solubility rules, and predict and analyse the composition of substances when two ionic solutions are mixed for example: - Potassium chloride and silver nitrate - Potassium iodide and lead nitrate - Sodium sulfate and barium nitrate Solubility Rules: - To work out if a precipitate forms, the K sp values from the data table can be used. Alternatively, we can use the solubility rules. - Insoluble and slightly soluble mean that a precipitate will form, and ions will form in most cases - Soluble means no precipitate will form, and the ions will remain solvated.

Predicting the formation of a Precipitate using K sp : - The formation of a precipitate is dependent on the quantities of the two solutions mixed together. Example: Consider the equilibrium in a saturated solution of barium hydroxide,

−¿ ¿ 2+¿+2 OH (aq) ¿ Ba ( OH )2(s) ⇋ Ba (aq ) 2+¿ Ba¿ ¿ −¿ OH ¿ K sp =¿ -

The solubility constant for this at 25°C is 0.000255.

-

-

-

2+¿ Ba¿ ¿ This means −¿ in a saturate solution. ¿ OH ¿ ¿ 2+¿ −¿ If the concentration of and OH ¿ are higher than the amounts in a saturated solution, Q > Ba¿ K sp , precipitation occurs. 2+¿ −¿ and OH ¿ are lower than the amounts in a saturated solution, Q < K sp If the concentration of Ba¿ , precipitation does not occur.

Derive equilibrium expressions for saturated solutions in terms of substance from its K sp value.

K sp

and calculate the solubility of an ionic

Equilibria in Saturated Solutions: - When a solvent has dissolved all the solute it can at a given temperature, the resulting solution is saturated. - In a saturated solution, the system is at equilibrium. - The forward and reverse reaction have the same rate. In other words, when an ion dissolves, another precipitates simultaneously. - The equilibrium constant for these solution equilibria is called the solubility product constant. It is given the given the symbol ‘ K sp ’. - The equilibrium constant is for the equation written in the direction of dissolution. - The solubility product constant characterises the solubility of a substance. Example: Write the dissolution equation and the equilibrium expression for

−¿ ¿ 3+ ¿ + 3 OH (aq) Al ( OH ) 3l → Al¿(aq)

K sp , for aluminium hydroxide,

Al ( OH )3 .

3+¿ Al¿ ¿ −¿ ¿ OH ¿ ¿ K sp =¿ Example: What will happen if more water is added to a saturated solution of CaSO 4 . - There will be a decrease in the concentration of ions as the volume has been increased, the equilibrium will shift to the right (Le Chatelier’s Principle), to form a new equilibrium, K will stay the same therefore solid dissolves until same concentrations have been achieved or no solid left. Example: What will happen to the concentrations of the ions and hence the position of the equilibrium if more solid solute is added to a saturated solution of CaSO 4 . - Adding more solute to an already saturated solution will have zero effect and will not disturb the equilibrium as saturated solution means that no more solute can be dissolved in the given volume. Thus, the concentration of the solution stays the same, rate of forward and reverse reactions are same and K does not change. Note: - Do not compare the relative solubility of salts directly from the table of K sp values if they do not produce the same total number of ions. - For comparison of solubility of salts that produce a different total number of ions, the solubility needs to be calculated.

Calculating K sp from Solubility: - To perform calculations involving moles per litre.

-

K sp , we need to know the concentrations of the ions in the solution in

Unless the density is given assume 1Kg is equivalent of 1 litre of solution.

Example: The solubility of sodium sulfate in water at 20°C is 139 g/L. Calculate the

K sp , at this temperature.

Step 1: Write a balanced equation for the dissolution of the compound, and the corresponding equilibrium expression.

2−¿ ¿ +¿+ SO (aq ) ¿ Na2 SO 4( s) → 2 Na(aq) Step 2: Convert the given solubility to mol/L.

139

[ Na2 SO4(s) ]=134 g L−1 = 2 ( 22.99 ) +32.07+4( 16) =0.978 molL−1 Step 3: Use the mole ratio in the balanced equation to calculate the concentration of the cation and anion in the saturated solution.

+¿¿ Na ¿ ¿ 2−¿ SO 4¿ ¿ ¿

Step 4: Substitute the concentrations into the equilibrium constant expression. Express your answer to the correct number of significant figures.

K sp= [ 1.95] × [ 0.978 ] 2

K sp =3.747

K sp=3.75 molL

−1

Calculating Solubility from K sp : - Solubility is usually expressed as moles per litre or mass per volume. Example: −5 The solubility constant for calcium sulfate at 25°C is 4.93 ×10 . Calculate its solubility at this temperature in mol/L. Step 1: Write an equation for the dissolution of the compound and the corresponding equilibrium expression.

( aq ) SO2−¿ ¿ 2+¿+ ¿ ¿ CaSO 4( s) → Ca( aq )

2+¿ Ca¿( aq ) ¿ ( aq) 2−¿ SO ¿ ¿=4.93 ×10−5 K sp =¿

Step 2: Let the moles of the ionic compound that dissolve in 1L be ‘ x ’. Hence, calculate the concentrations of the cation and anion at equilibrium.

[CaSO 4( s) ] =x mol

2+ ¿ Ca¿(aq ) ¿ Step 3: Substitute the pronumerals and the given value of ( aq) 2−¿ SO ¿ ¿= x mol ¿

K sp into the

equilibrium expression. Solve for x, then give your answer to the correct number of significant figures, with the correct units.

2+¿ Ca¿( aq ) ¿ ( aq) 2−¿ SO ¿ ¿=4.93 ×10−5 K sp =¿ -

-

−5

4.93 ×10 =x

2

−1

x=7.02 molL

The Common Ion Effect:

In an aqueous solution of an ionic compound, the ions are dissociated. This means that the ions separate into individual solvated ions and are indistinguishable from their origin. In a saturated solution, if another substance is added that has an ion in common with the first substance, it will affect the position of the equilibrium, leading to a lower solubility. This is known as the common ion effect. The common ion effect can be also predicted using Le Chatelier’s principle and collision theory.

Example: Consider the dissolution of barium hydroxide. - If hydroxide ions (in the form of sodium hydroxide) are added to the solution, then the concentration of hydroxide will increase. - Due to K sp remaining constant at constant temperature, the concentration of Barium ions will have to decrease. The concentration of barium can decrease if it combines with the Hydroxide ions to form solid Barium Hydroxide. - The result is a shift in the equilibrium towards the left hand side, so less barium hydroxide will be dissolved. Example: Explain the formation of more solid Barium Hydroxide when Sodium Hydroxide powder is added to a saturated solution of Barium Hydroxide, in terms of Le Chatelier’s principle. - When Sodium hydroxide is added, it will dissociate into Sodium and Hydroxide ions, increasing the overall concentration of the hydroxide ions in the solution. This change in concentration will disturb the equilibrium and shift it to the left due to Le Chatelier’s principle to consume the excess. This also consumes the barium hydroxide ions and decreases the overall concentration of the Barium ions. A solid precipitate of Barium Hydroxide forms out of the solution leaving less Barium Hydroxide dissolved in the solution. Example: Explain the formation of more solid barium hydroxide when sodium hydroxide powder is added to a saturated solution of barium hydroxide, in terms of collision theory. - Sodium hydroxide dissolves to form sodium ions and hydroxide ions. As there is a higher concentration of the hydroxide ions, there are more collisions. Hence, the rate of the reverse reaction increases relative to the forward reaction and equilibrium shifts to the left to produce more solid Barium Hydroxide.

Calculations involving the Common Ion Effect: - ICE calculations can be used for common ion calculations, with the initial concentration of the common ion being non-zero. Example: What is the solubility of Lead(ll) Chloride, in mol/L, in a 0.15M Sodium Chloride solution? Step 1: Write an equation for the dissolution of the compound and the corresponding equilibrium expression.

−¿ 2 + ¿+ 2 Cl ¿(aq ) ¿ PbCl2(s ) ⇋ Pb(aq)

Step 2: Construct an ICE table - Calculate the concentration of the common ion initially. - Let the concentration of the dissolved ionic compound be ‘x’. - Hence, calculate the concentrations of the ions at equilibrium.

−¿¿ 2Cl (aq )

2+ ¿ ¿ Pb(aq)

Concentration (mols) Initial Change Equilibrium

0 +x x

0.15 +2x 0.15 + 2x

Step 3: Assume that since K sp is small, dissolution does not change the concentration of the common ion. Substitute the pronumerals and the given value of K sp into the equilibrium expression and solve for ‘x’.

K sp =x ( 0.15+2 x)

K sp =x [ 0.15]

2

−5

K sp =1.7 × 10

−5

x=

1.7× 10 0.15

x=7.6 ×10

−4

Step 4: Check the assumption (moles of common ion added by dissolution...