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# Strong Acid-Weak Base and Weak Acid - Strong Base

Strong Acid-Weak Base and Weak Acid - Strong Base. CH 3 COOH(aq) + OH - (aq) -&gt; CH 3 COO - (aq) + H 2 O(l). 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

## Strong Acid-Weak Base and Weak Acid - Strong Base

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1. Strong Acid-Weak Base and Weak Acid - Strong Base

2. CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l) Slow change in pH before equivalence point; solution is a buffer CH3COOH(aq)/CH3COO-(aq) At halfway point [HA] = [A-] pH = pKa At equivalence, pH determined by CH3COO-(aq)

3. Changes in pH during a titration of a weak acid/base with a strong base/acid: Halfway to the stoichiometric point, the pH = pKa 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

4. Titration of 100.0 mL of 0.1000 M CH3COOH(aq) with 0.1000 M NaOH Before addition of NaOH: pH determined by CH3COOH(aq) CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq) Answer: pH = 2.88 Before the equivalence point: determine pH for a buffer Addition of 30.00 mL of NaOH(aq) The OH-(aq) reacts with the CH3COOH(aq). Determine concentration of CH3COOH(aq) and CH3COO- (aq) in solution after addition of the base. Answer: pH = 4.38 At half equivalence: [CH3COOH(aq)] = [CH3COO-(aq)] pH = pKa

5. At equivalence: enough OH-(aq) added to react with all CH3COOH(aq). For this problem, equivalence is reached when 100.0mL of OH- is added; i.e. 0.01000 moles of OH-(aq) added Solution contains 0.01000 moles CH3COO-(aq) in 200.0 mL solution; [CH3COO-(aq)] = 0.05000 M pH determined by CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH- (aq) pH = 8.72 (note greater than 7.0) Beyond equivalence: pH determined by excess OH-(aq)

6. Estimate the pH at the equivalence point of the titration of 25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq) (Ka(HCOOH) = 1.8 x 10-4) At the equivalence point, enough NaOH(aq) has been added to react with all the HCOOH(aq) forming HCOO-(aq) The reaction: HCOO-(aq) + H2O(l)  HCOOH(aq) + OH- determines the pH at equivalence Answer: 8.26

7. Polyprotic acid and bases Titration of H2CO3 with a strong base

8. Indicators A compound whose color changes noticeably over a short range of pH. pH = 7.0 8.5 9.4 9.8 12.0

9. [In- (aq)] [H3O+(aq)] KIn = [HIn(aq) ] The indicator is a weak acid itself (HIn) HIn(aq) + H2O(l)  In-(aq) + H3O+ (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 = pKIn

10. In choosing an indicator: pKIn ≈ pH(equivalence point) ± 1

11. Applications Atmospheric CO2(g) CO2(g) + H2O(l)  H2CO3(aq) H2CO3(aq) + H2O(l)  HCO3-(aq) + H3O+(aq) Acid rain: When pH < 5.5 due to pollutants in the air like SO2, SO3, NO2, 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

12. 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 (H2PO4-/HPO42-) H2PO4-(aq) + H2O(l)  H3O+(aq) + HPO42-(aq) Ka2 = 6.2 x 10-8, pKa2 = 7.21 Average pH of blood is 7.40 indicates that [HPO42-(aq)] / [H2PO4-(aq) ] = 1.55

13. 2) HCO3-/H2CO3 buffer CO2(g) + H2O(l)  H2CO3(aq) H2CO3(aq) + H2O(l)  HCO3-(aq) + H3O+(aq) pKa1 = 6.36 [HCO3-(aq)] / [H2CO3(aq) ] = 11.0 Build up of H2CO3 would destroy this balance. In the body H2CO3(aq)  H2O(l) + CO2 (g) CO2 is exhaled from the lungs to prevent buildup of H2CO3

14. 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 CaCO3 dissolve to a very small extent in water - sparingly soluble

15. The Earth’s crust consists largely of sparingly soluble salts; e.g. gypsum (CaSO4.2H2O), calcite (CaCO3), dolomite (xCaCO3.yMgCO3), oxides and sulfides of metals such as Fe, aluminosilicates (XAlSi3O8 or XAlSi2O8, X = Na+, K+, Ca2+) Hard-water contains high levels of Ca2+ and Mg2+ Ca2+ forms soap scum with detergents. Add soluble Na2CO3 (washing soda) to precipitate CaCO3 which washes off. Chemical weathering includes the dissolving of sparingly soluble salts. CaCO3(s) + CO2(g) + H2O(l)  Ca2+(aq) + 2 HCO3-(aq)

16. Highly soluble compounds: several grams of the compound dissolves per 100 g of water At 298 K: 36 g of NaCl, 122 g of AgNO3 Sparingly soluble compounds: less than one gram dissolves per 100 g of water At 298 K: 2.4 x 10-4 g of AgCl; 9.3 x 10-4 g of CaCO3, 4.4 x 10-14 g of PbS

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

18. The molar solubility of Ag2Cr2O4 is 6.5 x 10-5 mol/L at 25o C. Determine the value of Ksp. Ag2Cr2O4(s)  2 Ag+ (aq) + Cr2O42- (aq) Ksp = [Ag+ (aq) ]2 [Cr2O42- (aq)] [Ag+ (aq) ] = 2 x 6.5 x 10-5 mol/L [Cr2O42- (aq) ] = 6.5 x 10-5 mol/L Ksp = 1.1 x 10-12

19. Determine the solubility of BaSO4(s) in pure water at 298 K in moles/liter and grams/liter. Ksp (BaSO4) = 1.1 x 10-10 BaSO4(s)  Ba2+ (aq) + SO42- (aq) If x is the solubility in moles/liter [Ba2+(aq)] [SO42-(aq)] = x2 = 1.1 x 10-10 [Ba2+(aq)] = [SO42-(aq)] = 1.0 x 10-5 M Solubility of BaSO4 is 1.0 x 10-5 M or 2.3 x 10-3 g/L

20. Precipitation from Solution If equal volumes of aqueous solutions of 0.2 M Pb(NO3)2 and KI are mixed will PbI2(s) precipitate out? Ksp of PbI2 is 1.4 x 10-8 Use the reaction quotient, Q, to predict whether precipitation will occur Pb(NO3)2 (aq) + KI(aq) -> PbI2 (s) + KNO3 (aq) Net ionic equation: Pb2+ (aq) + 2I- (aq) -> PbI2 (s) The reverse of this reaction defines Ksp PbI2 (s)  Pb2+ (aq) + 2I- (aq)

21. Ksp = [Pb2+ (aq)] [I- (aq)]2 If Q > Ksp precipitation; if Q < Ksp no precipitation Equal volumes of Pb(NO3)2 and KI are mixed On mixing, volume of mixed solution is twice initial volume [Pb2+ (aq)] = 0.2M / 2 = 0.1 M [I- (aq)] = 0.1 M Q = [Pb2+(aq)] [I- (aq)]2 = (0.1)(0.1)2 = 0.001 M Q > Ksp; PbI2(s) precipitates

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