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 Effect Weak acid: HA + H 2 O ⇌ H 3 O + + 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? 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 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 Consist of a weak conjugate acid-base pair Control pH at a desired level (pK a ) Control pH at a desired level (pK a ) Examples: blood (p. 713), physiological fluids, seawater (p. 728), foods Examples: blood (p. 713), physiological fluids, seawater (p. 728), foods

How do buffers work?

Buffer Calculations

Calculating the pH of a Buffer 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: H 3 O + + HA or A - → Adding base: OH - + HA or A - → Calculating pH: 1. Stoichiometry of added acid or base 2. 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 NH 3 and 0.15 M NH 4 Br. K b =1.8x10 -5 for NH 3

Calculate the pH for a 1.0-L solution of 0.25 M NH 3 and 0.15 M NH 4 Br after the addition of 0.05 mol of RbOH.

Calculate the pH for a 1.0-L solution of 0.25 M NH 3 and 0.15 M NH 4 Br after the addition of 0.35 mol of HCl.

Buffers (wrap up) H-H equation H-H equation No 5% check No 5% check When a strong acid or base is added, start the reaction with that acid or base. 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 Making buffers of a specific pH? H-H equation Buffer capacity exceeded – when added acid or base totally consumes a buffer component (p. 726) 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? K a = 1.3x for phenol

17.3 Acid-Base Titrations Titration – a reaction used to determine concentration (acid-base, redox, precipitation) Titration – a reaction used to determine concentration (acid-base, redox, precipitation) Titrant – solution in buret; usually a strong base or acid Titrant – solution in buret; usually a strong base or acid Analyte – solution being titrated; often the unknown Analyte – solution being titrated; often the equivalence point (or stoichiometric point): mol acid = mol equivalence point (or stoichiometric point): mol acid = mol base Found by titration with an indicator (pp ) Found by titration with an indicator (pp ) Solution not necessarily neutral Solution not necessarily neutral pH dependent upon salt formed pH dependent upon salt formed pH titration curve – plot of pH vs. titrant volume pH titration curve – plot of pH vs. titrant volume

Acid-base Titration Reactions and Curves TypeAcidBase 1strongstrong 2weakstrong 3strongweak Recognize curve types Recognize curve types Calculate pH at various points on curve. Calculate pH at various points on curve.

Type 1: Strong acid + strong base Goes to completion: H 3 O + + OH - → 2 H 2 O Goes to completion: H 3 O + + OH - → 2 H 2 O Forms a neutral salt Forms a neutral salt Equivalence point - neutral solution, [H 3 O + ] = 1.0 x M, pH = 7.00 Equivalence point - neutral solution, [H 3 O + ] = 1.0 x M, pH = 7.00 pH calculations involve only stoichiometry and excess H 3 O + or OH - pH calculations involve only stoichiometry and excess H 3 O + or OH -

Strong acid – Strong base

Type 1: Strong acid + strong base 20.0 mL M HClO 4 titrated with M KOH mL base mmol base added mmol acid remain total mL [H 3 O + ] pH Initial mmol acid =

Another SB/SA titration 10.0 mL 0.20 M KOH titrated with 0.10 M HCl mL acid mmol acid added mol base remain total mL [OH - ] pH Initial mmol base =

Type 2: Weak acid + strong base Titration reaction goes to completion: Titration reaction goes to completion: HA + OH - → A - + H 2 O HA + OH - → A - + H 2 O Forms a basic salt (from conj. base of the weak acid) Forms a basic salt (from conj. base of the weak acid) Equivalence point - basic solution, pH > 7.00 Equivalence point - basic solution, pH > 7.00 pH calculations involve stoichiometry and equilibrium pH calculations involve stoichiometry and equilibrium

Weak acid – Strong base

Effect of K a on Titration curve

Polyprotic acid – Strong base

Type 2: Weak acid + strong base 25.0 mL 0.100M HC 3 H 5 O 2 titrated with M KOH K a = 1.3x10 -5 Calculate the pH at the following points: A. Initial (0.00 mL KOH) B mL KOH C. Midpoint (12.50 mL KOH) D. Equivalence pt. (25.00 mL KOH) E mL after eq. pt. (35.00 mL KOH)

Polyprotic Weak acid – Strong base

Type 3: Weak base + strong acid Titration reaction goes to completion: Titration reaction goes to completion: H 3 O + + B → BH + + H 2 O H 3 O + + B → BH + + H 2 O Forms an acidic salt (from conj. acid of the weak base) Forms an acidic salt (from conj. acid of the weak base) Equivalence point - acidic solution, pH < 7.00 Equivalence point - acidic solution, pH < 7.00 pH calculations involve stoichiometry and equilibrium pH calculations involve stoichiometry and equilibrium

Strong base Weak base Strong base – Strong acid Weak base – Strong acid

Type 3: Weak base + strong acid 25.0 mL M NH 3 titrated with M HCl K b = 1.8x10 -5 Calculate the pH at the following points: A. Initial (0.00 mL HCl) B. Midpoint (______ mL HCl) C mL HCl D. Equivalence pt. (______ mL HCl) E mL after eq. pt. (______ mL HCl)

Types 2 & 3 pH Calculation Summary Initial pH – same as weak acid or base problem (chapter 16) Initial pH – same as weak acid or base problem (chapter 16) Before equivalence point – Buffer Before equivalence point – midpoint – half of the weak analyte has been midpoint – half of the weak analyte has been neutralized [weak acid] = [conj. base] or [weak base] = [conj. acid] [H 3 O + ] = K a and pH = pK equivalence point: mol acid = mol base; equilibrium problem with equivalence point: mol acid = mol base; equilibrium problem with conjugate Beyond equivalence point – pH based on excess titrant; stoichiometry 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) 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 Insoluble compound – compound with a solubility less than 0.01 M; also sparingly soluble Solubility rules are given on p. 121 (ch. 4) Solubility rules are given on p. 121 (ch. 4) Dissolution reaches equilibrium in water between undissolved solid and hydrated ions Dissolution reaches equilibrium in water between undissolved solid and hydrated ions

Solubility Product Constant, K sp Equilibrium constant for insoluble compounds Equilibrium constant for insoluble compounds Solid salt nor water included in expression Solid salt nor water included in expression Appendix D, p for values Appendix D, p for values BaSO 4 (s) ⇌ Ba 2+ (aq) + SO 4 2- (aq) PbCl 2 (s) ⇌ Pb 2+ (aq) + 2 Cl - (aq)

Solubility Product Calculations In concentration tables, x = solubility In concentration tables, x = solubility Problem types Problem types 1. solubility → K sp 2. K sp → solubility (estimate, see p. 726)

Comparing Salt Solubilities Generally: solubility ↑ K sp ↑ Can only compare K sp values if the salts produce the same number of ions Can only compare K sp values if the salts produce the same number of ions If different numbers of ions are produced, solubility must be compared. If different numbers of ions are produced, solubility must be compared.

17.5 Factors that Affect Solubility 1. Common-Ion Effect LeChatelier’s Principle revisited LeChatelier’s Principle revisited Addition of a product ion causes the solubility of the solid to decrease, but the K sp remains constant. 2. pH LeChatelier’s Principle again! LeChatelier’s Principle again! Basic salts are more soluble in acidic solution. Acidic salts are more soluble in basic solution. Environmental example: CaCO 3 – limestone Stalactites 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