Strong Acid-Weak Base and Weak Acid - Strong Base

Slow change in pH before equivalence point; solution is a buffer CH 3 COOH(aq)/CH 3 COO - (aq) At halfway point [HA] = [A - ] pH = pK a At equivalence, pH determined by CH 3 COO - (aq) CH 3 COOH(aq) + OH - (aq) -> CH 3 COO - (aq) + H 2 O(l)

Changes in pH during a titration of a weak acid/base with a strong base/acid: Halfway to the stoichiometric point, the pH = pK a of the acid The pH is greater than 7 at the equivalence point of the titration of a weak acid and strong base The pH is less that 7 at the equivalence point of the titration of a weak base and strong acid Beyond the equivalence point, the excess strong acid or base will determine the pH of the solution

Titration of mL of M CH 3 COOH(aq) with M NaOH Before addition of NaOH: pH determined by CH 3 COOH(aq) CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) Answer: pH = 2.88 Before the equivalence point: determine pH for a buffer Addition of mL of NaOH(aq) The OH - (aq) reacts with the CH 3 COOH(aq). Determine concentration of CH 3 COOH(aq) and CH 3 COO - (aq) in solution after addition of the base. Answer: pH = 4.38 At half equivalence: [CH 3 COOH(aq)] = [CH 3 COO - (aq)] pH = pKa

At equivalence: enough OH - (aq) added to react with all CH 3 COOH(aq). For this problem, equivalence is reached when 100.0mL of OH - is added; i.e moles of OH - (aq) added Solution contains moles CH 3 COO - (aq) in mL solution; [CH 3 COO - (aq)] = M pH determined by CH 3 COO - (aq) + H 2 O(l)  CH 3 COOH(aq) + OH - (aq) pH = 8.72(note greater than 7.0) Beyond equivalence: pH determined by excess OH - (aq)

Estimate the pH at the equivalence point of the titration of mL of M HCOOH(aq) with M NaOH(aq) (K a (HCOOH) = 1.8 x ) At the equivalence point, enough NaOH(aq) has been added to react with all the HCOOH(aq) forming HCOO - (aq) The reaction: HCOO - (aq) + H 2 O(l)  HCOOH(aq) + OH - determines the pH at equivalence Answer: 8.26

Polyprotic acid and bases Titration of H 2 CO 3 with a strong base

Indicators A compound whose color changes noticeably over a short range of pH. pH =

The indicator is a weak acid itself (HIn) HIn(aq) + H 2 O(l)  In - (aq) + H 3 O + (aq) K In = [HIn (aq) ] [H 3 O + (aq)][In - (aq)] The acid form, HIn, has a different color from the base form In - The end point of a titration is the point at which the concentrations of the acid and base forms of the indicator are equal; [HIn(aq)] = [In - (aq)] Color change occurs when pH = pK In

In choosing an indicator: pK In ≈ pH(equivalence point) ± 1

Applications Atmospheric CO 2 (g) CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)  HCO 3 - (aq) + H 3 O + (aq) Acid rain: When pH < 5.5 due to pollutants in the air like SO 2, SO 3, NO 2, which dissolve in water to form strong acids Many lakes have water too acidic to sustain life, forests have also been damaged. In the ground, acidic rain water can be neutralized by ions in the soil

The ocean is buffered to a pH of 8.4 by buffering that depends on the presence of hydrogen carbonates and silicates Physiological Buffers: Body fluids such as blood function over a very narrow pH range, maintained by buffers Blood contains two buffering systems: 1) Phosphate buffer (H 2 PO 4 - /HPO 4 2- ) H 2 PO 4 - (aq) + H 2 O(l)  H 3 O + (aq) + HPO 4 2- (aq) K a2 = 6.2 x 10 -8, pK a2 = 7.21 Average pH of blood is 7.40 indicates that [HPO 4 2- (aq)] / [H 2 PO 4 - (aq) ] = 1.55

2) HCO 3 - /H 2 CO 3 buffer CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)  HCO 3 - (aq) + H 3 O + (aq) pK a1 = 6.36 [HCO 3 - (aq)] / [H 2 CO 3 (aq) ] = 11.0 Build up of H 2 CO 3 would destroy this balance. In the body H 2 CO 3 (aq)  H 2 O(l) + CO 2 (g) CO 2 is exhaled from the lungs to prevent buildup of H 2 CO 3

Solubility Equilibria Many ionic solids dissociate into their ions in water: NaCl(aq) -> Na + (aq) + Cl - (aq). Compounds such as NaCl exist completely as Na + (aq) and Cl - (aq) in aqueous solutions unless the amount of NaCl exceeds the solubility of NaCl in water Other compounds such as CaCO 3 dissolve to a very small extent in water - sparingly soluble

The Earth’s crust consists largely of sparingly soluble salts; e.g. gypsum (CaSO 4.2H 2 O), calcite (CaCO 3 ), dolomite (xCaCO 3. yMgCO 3 ), oxides and sulfides of metals such as Fe, aluminosilicates (XAlSi 3 O 8 or XAlSi 2 O 8, X = Na +, K +, Ca 2+ ) Hard-water contains high levels of Ca 2+ and Mg 2+ Ca 2+ forms soap scum with detergents. Add soluble Na 2 CO 3 (washing soda) to precipitate CaCO 3 which washes off. Chemical weathering includes the dissolving of sparingly soluble salts. CaCO 3 (s) + CO 2 (g) + H 2 O(l)  Ca 2+ (aq) + 2 HCO 3 - (aq)

Highly soluble compounds: several grams of the compound dissolves per 100 g of water At 298 K: 36 g of NaCl, 122 g of AgNO 3 Sparingly soluble compounds: less than one gram dissolves per 100 g of water At 298 K: 2.4 x g of AgCl; 9.3 x g of CaCO 3, 4.4 x g of PbS

AgCl(s)  Ag + (aq) + Cl - (aq) Define an equilibrium constant for this process: solubility product, K sp K sp =[Ag + (aq)] [Cl - (aq)] The solubility product, K sp, is the equilibrium constant for the equilibrium between an undissolved salt and its ions in a saturated solution.

The molar solubility of Ag 2 Cr 2 O 4 is 6.5 x mol/L at 25 o C. Determine the value of K sp. Ag 2 Cr 2 O 4 (s)  2 Ag + (aq) + Cr 2 O 4 2- (aq) K sp = [Ag + (aq) ] 2 [Cr 2 O 4 2- (aq)] [Ag + (aq) ] = 2 x 6.5 x mol/L [Cr 2 O 4 2- (aq) ] = 6.5 x mol/L K sp = 1.1 x

Determine the solubility of BaSO 4 (s) in pure water at 298 K in moles/liter and grams/liter. K sp (BaSO 4 ) = 1.1 x BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) If x is the solubility in moles/liter [Ba 2+ (aq)] [SO 4 2- (aq)] = x 2 = 1.1 x [Ba 2+ (aq)] = [SO 4 2- (aq)] = 1.0 x M Solubility of BaSO 4 is 1.0 x M or 2.3 x g/L

Precipitation from Solution If equal volumes of aqueous solutions of 0.2 M Pb(NO 3 ) 2 and KI are mixed will PbI 2 (s) precipitate out? K sp of PbI 2 is 1.4 x Use the reaction quotient, Q, to predict whether precipitation will occur Pb(NO 3 ) 2 (aq) + KI(aq) -> PbI 2 (s) + KNO 3 (aq) Net ionic equation: Pb 2+ (aq) + 2I - (aq) -> PbI 2 (s) The reverse of this reaction defines K sp PbI 2 (s)  Pb 2+ (aq) + 2I - (aq)

K sp = [Pb 2+ (aq)] [I - (aq)] 2 If Q > K sp precipitation; if Q < K sp no precipitation Equal volumes of Pb(NO 3 ) 2 and KI are mixed On mixing, volume of mixed solution is twice initial volume [Pb 2+ (aq)] = 0.2M / 2 = 0.1 M [I - (aq)] = 0.1 M Q = [Pb 2+ (aq)] [I - (aq)] 2 = (0.1)(0.1) 2 = M Q > K sp ; PbI 2 (s) precipitates