Additional Aspects of Aqueous Equilibria
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Transcript of Additional Aspects of Aqueous Equilibria
Additional Aspects of Aqueous Equilibria
BLB 12th Chapter 17
Buffered Solutions (sections 1-2)Acid/Base Reactions & Titration Curves
(3)Solubility Equilibria (sections 4-5)
Two important points:1. Reactions with strong acids or
strong bases go to completion.2. Reactions with only weak acids and
bases reach an equilibrium.
17.1 The Common Ion EffectWeak acid:HA + H2O ⇌ H3O+ + A-
+Salt of conj. Base:NaA → Na+(aq) + A-(aq)
= two sources of A- Common Ion! What affect does the addition of its conjugate
base have on the weak acid equilibrium? On the pH?
Used in making buffered solutions
Calculate the pH of a solution containing 1.0 M HF and 0.60 M KF. (Recall that the pH of 1.00 HF is 1.58.)
Addition of F- shifts the equilibrium, reducing the [H+].
17.2 Buffered Solutions Resist a change in pH upon the
addition of small amounts of strong acid or strong base
Consist of a weak conjugate acid-base pair
Control pH at a desired level (pKa) Examples: blood (p. 713),
physiological fluids, seawater (p. 728), foods
How do buffers work?
Buffer Calculations
Calculating the pH of a Buffer
][][log
][][loglog]log[
][][][
3
3
acidbasepKpH
AHAKOH
AHAKOH
a
a
a
Henderson-Hasselbalch equation
Calculate the pH of a solution containing 1.0 M HF and 0.60 M KF. (again, but the easy way)
Addition of Strong Acids or Bases to Buffers
Adding acid: H3O+ + HA or A- →
Adding base: OH- + HA or A- →
Calculating pH:1. Stoichiometry of added acid or base2. Equilibrium problem (H-H equation)
Calculate the pH after adding 0.20 mol of HCl to 1.0 L of the 1.0 M HF and 0.60 M KF buffer.
Calculate the pH after adding 0.10 mol of NaOH to 1.0 L of the 1.0 M HF and 0.60 M KF buffer.
Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br. Kb=1.8x10-5 for NH3
Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.05 mol of RbOH.
Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.35 mol of HCl.
Buffers (wrap up) H-H equation No 5% check When a strong acid or base is
added, start the reaction with that acid or base.
Making buffers of a specific pH? H-H equation
Buffer capacity exceeded – when added acid or base totally consumes a buffer component (p. 726)
How would you prepare a phenol buffer to control pH at 9.50? Ka = 1.3x10-10 for phenol
17.3 Acid-Base Titrations Titration – a reaction used to determine
concentration (acid-base, redox, precipitation) Titrant – solution in buret; usually a strong base
or acid Analyte – solution being titrated; often the
unknown @ equivalence point (or stoichiometric point):
mol acid = mol base Found by titration with an indicator (pp. 721-722) Solution not necessarily neutral pH dependent upon salt formed
pH titration curve – plot of pH vs. titrant volume
Acid-base Titration Reactions and Curves
Type Acid Base
1 strong strong
2 weak strong
3 strong weak
Recognize curve types
Calculate pH at various points on curve.
Type 1: Strong acid + strong base Goes to completion: H3O+ + OH- → 2
H2O Forms a neutral salt Equivalence point - neutral solution,
[H3O+] = 1.0 x 10-7 M, pH = 7.00 pH calculations involve only
stoichiometry and excess H3O+ or OH-
Strong acid – Strong base
Type 1: Strong acid + strong base20.0 mL 0.200 M HClO4 titrated with 0.200 M KOH
mL base
mmol base
added
mmol acid
remain
total mL
[H3O+] pH
0.0010.0020.0030.0040.00
Initial mmol acid =
Another SB/SA titration10.0 mL 0.20 M KOH titrated with 0.10 M HCl
mL acid
mmol acid added
mol base
remain
total mL
[OH-] pH
0.0015.0020.0035.0050.00
Initial mmol base =
Type 2: Weak acid + strong base Titration reaction goes to completion:
HA + OH- → A- + H2O Forms a basic salt (from conj. base of
the weak acid) Equivalence point - basic solution, pH >
7.00 pH calculations involve stoichiometry
and equilibrium
Weak acid – Strong base
Effect of Ka onTitration curve
Polyprotic acid – Strong base
Type 2: Weak acid + strong base25.0 mL 0.100M HC3H5O2 titrated with 0.100 M KOHKa = 1.3x10-5
Calculate the pH at the following points:
A. Initial (0.00 mL KOH)B. 10.00 mL KOHC. Midpoint (12.50 mL KOH)D. Equivalence pt. (25.00 mL KOH)E. 10.00 mL after eq. pt. (35.00 mL
KOH)
Polyprotic Weak acid – Strong base
Type 3: Weak base + strong acid Titration reaction goes to completion:
H3O+ + B → BH+ + H2O Forms an acidic salt (from conj. acid of
the weak base) Equivalence point - acidic solution, pH <
7.00 pH calculations involve stoichiometry
and equilibrium
Strong base
Weak base
Strong base – Strong acidWeak base – Strong acid
Type 3: Weak base + strong acid 25.0 mL 0.150 M NH3 titrated with 0.100 M HCl Kb = 1.8x10-5
Calculate the pH at the following points:
A. Initial (0.00 mL HCl)B. Midpoint (______ mL HCl)C. 25.00 mL HClD. Equivalence pt. (______ mL HCl)E. 10.00 mL after eq. pt. (______ mL
HCl)
Types 2 & 3 pH Calculation Summary
Initial pH – same as weak acid or base problem (chapter 16)
Before equivalence point – Buffer @ midpoint – half of the weak analyte has
been neutralized [weak acid] = [conj. base] or [weak base] = [conj.
acid] [H3O+] = Ka and pH = pKa
@ equivalence point: mol acid = mol base; equilibrium problem with conjugate
Beyond equivalence point – pH based on excess titrant; stoichiometry
Test #2 Summary for Acid/Base problems
1. Weak acid or weak base only (ch. 16)
2. Buffer
3. SA + SB Titration
4. WA + SB or WB + SA Titration
17.4 Solubility Equilibria Solubility – maximum amount of
material that can dissolve in a given amount of solvent at a given temperature; units of g/100 g or M (ch. 13)
Insoluble compound – compound with a solubility less than 0.01 M; also sparingly soluble Solubility rules are given on p. 121 (ch.
4) Dissolution reaches equilibrium in water
between undissolved solid and hydrated ions
Solubility Product Constant, Ksp Equilibrium constant for insoluble
compounds Solid salt nor water included in
expression Appendix D, p. 1063 for values
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)
Solubility Product Calculations In concentration tables, x =
solubility Problem types
1. solubility → Ksp
2. Ksp → solubility (estimate, see p. 726)
Comparing Salt SolubilitiesGenerally: solubility ↑ Ksp ↑
Can only compare Ksp values if the salts produce the same number of ions
If different numbers of ions are produced, solubility must be compared.
17.5 Factors that Affect Solubility1. Common-Ion Effect
LeChatelier’s Principle revisitedAddition of a product ion causes the solubility of the solid to decrease, but the Ksp remains constant.
2. pH LeChatelier’s Principle again!
Basic salts are more soluble in acidic solution.Acidic salts are more soluble in basic solution.
Environmental example: CaCO3 – limestoneStalactites and stalagmites form due to changing pH in the water and, thus, the solubility of the limestone. (p. 948)
Common-ion Effect
Effect of pH on Solubility