aqueous solutions and solubility equilibria
DESCRIPTION
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 ) + H 2 O(l). - PowerPoint PPT PresentationTRANSCRIPT
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.