AP Chemistry Chapter 17

The Common-Ion Effect

- "A weak electrolyte will ionize less when a strong electrolyte with a common ion is added to the solution."
- Raises the pH

Buffered Solutions

- Buffers are solutions of a weak conjugate acid-base pair.
- They are particularly resistant to pH changes, even when strong acid or base is added.
- A buffer resists changes in pH because it contains both an acid to neutralize OH- ions and a base to neu

Making buffers

1) Buffers are usually made by mixing a weak acid or a weak base with a salt with the conjugate of that acid or base. (Example: Acetic acid with sodium acetate or Ammonia with ammonium chloride)
2) Can also be made by partial neutralization of a weak acid

How buffers work

- Here is a buffer system consisting of a weak acid and one of its salts (MX):
HX -> H+ + X-
Its acid-dissociation constant is
Ka = [H+][X-] / [HX]
Solving this expression for [H+], we have
[H+] = Ka [HX] / [H-]

In a buffer system [H+] is on....

The [H+] (and pH) are then based on 2 factors:
1) The value of the weak acids Ka
2) Ratio of the acid / base pair [HX] / [X]
- When a small amount of OH- is added to the buffer, the OH- reacts with HX to produce X- and water. But the [HX] / [X-] ratio rem

Calculating the pH of a buffer

Consider the equilibrium expression for the dissociation of a generic acid, HA.
HA + H2O -> H3O + A-
Ka = [H30+][A-] / [HA]
which becomes: Ka = [H30+] x [A-]/[HA]
Taking the negative log of both sides, we get
- log Ka = -log [H30+] + (- log [A-] / [HA] )

Henderson- Hasselbalch Equation

If [base] = [acid] or equimolar, then pH = pKa
If [base] < [acid] then pH < pKA
If [base] > [acid] then pH > pKa

pH range

- Range of pH values over which a buffer system works effectively
- Buffers best resist change in pH in either direction if the weak acid and its conjugate base are equimolar, meaning pH = pKa. Because the buffering action becomes poor once the concentrat

Finding the pH after adding an acid or base to a buffer stystem

1) Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
2) Use the Henderson- Hasselbalch equation to determine the new pH of the solution

Acid-Base titrations

- In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base)
- A pH meter or indicator are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equ

Titrations of a Strong Acid with a Strong Base

1) At the start, the pH is determined by the [acid].
pH = -log[acid]
2) after the start of the titration is near the equivalence point, the pH goes up slowly.
[H+] = mol H+ - mol OH - / total volume
pH = -log [H+]
3) At the equivalence point, moles acid =

Information about Titrations of a weak acid with a strong base

- Unlike with a titration of a strong acid with a strong base, the conjugate base of the acid affects the pH when it is formed.
- At the equivalence point the pH is > 7
- Use Phenolphthalien indicator

Titrations of a weak acid with a strong base

1) At the start, the pH is determined by the concentration of the [weak] acid and Ka.
2) Between the initial pH and the equivalence point, a partial neutralization creates a buffer solution so you have to find the [HA] and [H-] and use Henderson-Hasselbal

Weak acid with a strong base

1) No OH-; start
2) H+ > OH- : before equivalence
3) H+ = OH- : equivalence point
4) H+ < OH- after equivalence point
- With weaker acids the pH is higher and pH changes near the equivalence point are more subtle

Weak base with a strong acid

- pH < 7
- 1) No H+; start
2) H+ < OH -; before equivalence
3) H+ = OH-: equivalence point
4) H+ > OH-: after equivalence point
- When one titrates a polyprotic acid with a base, there in an equivalence point for each dissocation.

Solubility Equilibria

-Consider the equilibrium that exists is a saturated solution of BaSO4 in water:
BaSO4 -> Ba 2+ + SO4^2-
- For a saturated solution, the rate of dissolving and the rate of crystallization are equal.
- The equilibrium constant expression for this equilibri

Factors the Affect Solubility

1) Presence of common ions
2) the pH of the solution
3) the presence of complexing gents

The Common Ion effect

- If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt with decrease

Solubility and pH

- If a substance has a basic anion, it will be more soluble in an acidic solution (Example: Mg(OH)2 -> Mg 2 + 2OH-)
- Substances with acidic cations are more soluble in basic solutions

Formation of Complex Ions

- Metal ions can bond to a group of surrounding molecules or ions to form a complex ion.
- The number of Lewis bases that bond t the metal is often twice the metal ions charge!
- Metal ions can act as Lewis acids and form complex ions with Lewis bases (ca

Amphoterism

Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
- Examples are cation like Al3+, Zn 2+, Sn 2+
- Al(OH)3 -> Al3+ + 3OH_
- Al9OH03 + 3OH -> Al(OH04

Precipitation and separation of ions

- When ions are mixed, the reaction quotient (Q) can be used to determine the direction in which a reaction must proceed to reach equilibrium
- Will a precipitate form when ions of a low solubility are combined?
- If Q = K the system is at equilibrium and

Analysis of metal ions in group 1

- When HCL is added (H- and Cl-) only AgCl, Hg2Cl2, and PbCl2 have low enough solubility's to cause precipitation with chloride.
- All other cations stay dissolved

Analysis of metal ions in group 2

- After filtration, the solubility, now acidic, is treated with H2S.
- Any metal with a low solubility with sulfide (low Ksp) will precipitate
- Why is acidic important?
- S2- ( a basic anion) will react with the acid, shifting the solubility equilibrium

Analysis of metal ions in group 3

- After filtration, the solution is made basic and (NH4)2S is added.
- Why make it basic?
- In a basic solution, the sulfide ion increases [S2-] (remember acidic lowered the [S2-], so basic will raise it)
- Adding (NH4)2S causes the S2- concentration to i

Analysis of metal ions in group 4

- At this point, only group 1A and 2A ions are left
- Group 2A ions are insoluble with phosphate, so adding (NH3)2HPO4 causes the group 2A ions to precipitate

Analysis of metal ions in group 3

- Anything has to be an alkali metal or ammonium.
- A flame test of the original sample should show if any alkali metals are present.