Aqueous Solutions and Solubility Equilibria

Chapter 9

9.1 The Acid-Base Properties of Salt Solutions

• Salt: any ionic compound that is formed in a neutralization reaction from the anion of an acid and the cation of a base.

E.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

• The acid/base property of a salt results from reactions between water and the dissociated ions of the salt. – Ion that react with water produce a solution with

excess H30+ or OH

Salts that form Neutral Solutions

• Salts of strong bases and strong acids dissolve in water and form neutral solutions.– The conjugate bases of strong acids are very

___________.– The conjugate acids of strong bases are very

___________.E.g.

Neither ion will react with water: too weak!

Salts the Dissolve and form Acidic and Basic Solutions

• Salts of weak bases and strong acids dissolve in water and form acidic solutions.

• Salts of strong bases and weak acids dissolve in water and form basic solutions.

(LOOKING AT THE CONJUGATE ACIDS/BASES!)

Salts of Weak Acids and Bases

• Ions in a salt from weak acids and bases BOTH react with water.

• Acidity/basicity depends on relative strength of the ions.

• Determine which ion is stronger by comparing the Ka and Kb associated with the cation and anion.

• If Ka>Kb, acidic. If Kb>Ka, basic.

CLASSWORK/HOMEWORK

Read through SP on pg. 423. Make notes.Do PPs 1-3.

Calculating pH at Equivalence• Equivalence point: just enough acid and base have been

mixed for complete reaction to occur, with no excess of either reactant.

• Acid-base indicator: weak, monoprotic acid. It is in equilibrium between undissociated acid (one colour) and conjugate base (different colour)

• End-point: indicator changes colour.

PPs 8 & 9

SR (pg. 429) #2, 6, 7.

9.2 – Solubility Equilibria

Solubility as an Equilibrium Process

• Change in enthalpy, entropy, and temperature determine whether or not a change is favoured.

• Same is important to determine how much salt will dissolve.

• Change is favoured when G is negative. • When salt dissolves, entropy is increased.

Ex// Barium sulfate crystals in water:

As ions enter solution, rate of reverse change, recrystallisation, increases.

Eventually, rate of recrystallisation becomes equal to the rate of dissolving.

Solubility Equilibria

• Equilibrium exists between the solid ionic compound and its dissociated ions in solution.

Solubility Product Constant

Solubility product constant: Ksp!!!

• PPs 13, 15

Using the Solubility Product Constant

• Use the value of Ksp for a compound to determine the concentration of its ions in a saturated solution.

• Similar to finding equilibrium amounts using Kc for homogeneous equilibria.

WORK ON PPs 17, 18, 19,

The Common Ion Effect

• When ionic compound added to a solution that already contains one of its ions.

• Adding a common ion to a solution increases the concentration of that ion in solution – EQUILIBRIUM SHIFTS AAWY FROM THE ION. – Can form precipitates.

The Effect of a Common Ion on Solubility

• Work on PPs 21-23

9.3Predicting the Formation of a Precipitate

The Ion Product: Qsp: expression that is identical to the solubility product constant, but values are not necessarily at equilibrium.

(Where have we seen this before?)

E.g.

- Adding magnesium sulfate to water. - Initially, all magnesium sulfate dissolves. - Saturation: no more salt will dissolve. - More solid will form.

- Calculate Qsp by substituting the conc. Of each ion into the expression.

Using the Ion Product Expression

• If Qsp > Ksp, compound will form an ionic compound.

• How do we know which ionic compounds are soluble?

E.g.

• PPs 29, 30, 31

• PPs. 33, 34, 35

Test Review

Buffers and the Common Ion Effect

• A buffer consists of a ________________________ and a ________________________ or vise versa.

• Need to consider initial concentration of reactants since homogeneous.