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Calculating pH from K a

Calculating pH from K a. The K a for niacin (C 5 H 4 NCOOH) is 1.6 x 10 -5 . What is the pH of a 0.010 M solution of niacin?. C 5 H 4 NCOOH C 5 H 4 NCOO - ( aq ) + H + ( aq ) initial: 0.010 M 0 M 0 M change: - x M +x M +x M

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Calculating pH from K a

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  1. Calculating pH from Ka The Ka for niacin (C5H4NCOOH) is 1.6 x 10-5. What is the pH of a 0.010 M solution of niacin? C5H4NCOOH C5H4NCOO-(aq) + H+(aq) initial: 0.010 M 0 M 0 M change: - x M +x M +x M equilibrium: 0.010 - x M x M x M Ka = [C5H4NCOO-][H+] = x2 = 1.6 x 10-5 [C5H4NCOOH] (0.010 - x) x2 + 1.6 x 10-5x - 1.6 x 10-7 = 0 x = 3.9 x 10-4 pH = -log(3.9 x 10-4) = 3.41 For a small Ka like this one, you can simplify things by guessing that x<<0.010 and dropping the x from the denominator.

  2. Polyprotic Acids Polyprotic acids have more than one acid H. As a result, they have more than one Ka. H2CO3(aq) H+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7 HCO3- (aq) H+(aq) + CO32-(aq) Ka2 = 5.6 x 10-11 Note how small Ka2 is. It is always easier to remove the first proton from a polyprotic acid than the second. As long as successive Ka’s differ by a factor of 1000 or more, a good estimate of the pH of a polyprotic acid can be made from Ka1 alone.

  3. pH of Polyprotic Acids The solubility of carbon dioxide in water at 25°C is 0.0037 M. Assuming all of the carbon dioxide forms carbonic acid, find the pH of water saturated with carbon dioxide at 25°C. H2CO3(aq) H+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7 The larger Ka is, the stronger the acid is. 0.0037 M 0 M 0 M initial -x M +x M +x M change 0.0037 - x M x M x M equilibrium Ka1 = 4.3 x 10-7 = x2 assume x<<0.0037, x2 ≈ 1.591 x 10-9 (0.0037-x) x = 4.0 x 10-5 M pH = 4.40 4.0 x 10-5 is about 1% of 0.0037, which means our assumption that x<<0.0037 was valid.

  4. pH of Polyprotic Acids The solubility of carbon dioxide in water at 25°C is 0.0037 M. How many parts per billion (μg/L) carbonate is in water saturated with carbon dioxide at 25°C? HCO3-(aq) H+(aq) + CO32-(aq) Ka2 = 5.6 x 10-11 4.0 x 10-5 M 4.0 x 10-5 M 0 M initial -x M +x M +x M change (4.0 x 10-5 - x) (4.0 x 10-5 + x) x M equilibrium Ka2 = 5.6 x 10-11 = (4.0 x 10-5 + x)x assume x<< 4.0 x 10-5 (4.0 x 10-5-x) x = 5.6 x 10-11 Verify: 5.6 x 10-11 x 100 = .00014% < 5% 4.0 x 10-5

  5. pH of Polyprotic Acids The solubility of carbon dioxide in water at 25°C is 0.0037 M. How many parts per billion (μg/L) carbonate is in water saturated with carbon dioxide at 25°C? HCO3-(aq) H+(aq) + CO32-(aq) Ka2 = 5.6 x 10-11 4.0 x 10-5 M 4.0 x 10-5 M 0 M initial -x M +x M +x M change (4.0 x 10-5 – x) (4.0 x 10-5 + x) x M equilibrium x = 5.6 x 10-11 x = 5.6 x 10-11 mol/L x 60.01g/mol x 106μg/g = .0034 μg/L concentration of CO32- = 0.0034 ppb

  6. Weak Bases and the Base Dissociation Product Kb There are many organic bases that only partially ionize in water and therefore are weak bases. The calculation of pH of weak bases must incorporate the base-dissociation constant Kb NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = [NH4+][OH-] [NH3]

  7. Weak Bases and the Base Dissociation Product Kb Find the pH of a 0.15 M solution of ammonia. Kb(25°C) = 1.8 x 10-5 NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) 0.15 M 0 M 0 M initial -x M +x M +x M change 0.15-x M x M x M equilibrium Kb = [NH4+][OH-] = 1.8 x 10-5 = x2 (assume x<<0.15) [NH3] 0.15-x x = 1.6 x 10-3 M, so pOH = 2.78 pH = 11.22

  8. Recognizing Weak Bases Weak bases fall into two categories: 1. Compounds with a nonbonding pair of electrons that can serve as a proton acceptor. Most of these compounds contain nitrogen. -includes ammonia and the amines (hydroxylamine NH2OH and methylamine CH3NH2) 2. Anions of weak acids. - examples are ClO-, CO32-, HS-

  9. Ka and Kb of Conjugate Acid-Base Pairs NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = [NH4+][OH-] [NH3] NH4+(aq) NH3(aq) + H+(aq) Ka = [NH3][H+] [NH4+] Adding the equations gives H2O(l) H+(aq) + OH-(aq) KC = Ka x Kb but KC = Kw = KaKb ammonia acting as a base ammonium acting as an acid

  10. Ka and Kb of Conjugate Acid-Base Pairs Kw = KaKb This relationship shows that as the strength of an acid increases, the strength of its conjugate base decreases.

  11. Ka and Kb of Conjugate Acid-Base Pairs Kw = KaKb Because of this relationship, tables of data for weak acids normally do not show Kb. Another way to state this relationship is pKa + pKb = 14.00 at 25°C

  12. Ka and Kb of Conjugate Acid-Base Pairs Kw = KaKb What is the Ka at 25°C of NH4+, the conjugate acid of NH3? NH3 has a Kb of 1.8 x 10-5. NH4+ (aq) NH3(aq) + H+(aq) Ka = [NH3][H+] = Kw = 1.0 x 10-14 = 5.6 x 10-10 [NH4+] Kb 1.8 x 10-5

  13. Acid-Base Properties of Salt Solutions • Anions may be looked at as conjugate bases. • Cations may be looked at as conjugate acids. • The acid-base properties of salt solutions can be determined by the extent to which the ions in the solutions react with water.

  14. Acid-Base Properties of Salt Solutions 1. An anion that is the conjugate base of a strong acid (like Cl-) will not affect the pH of a solution. 2. An anion that is the conjugate base of a weak acid (like CN-) will increase the pH of a solution. CN-(aq) + H2O(l) HCN(aq) + OH-(aq)

  15. Acid-Base Properties of Salt Solutions 3. A cation that is the conjugate acid of a weak base (like NH4+) will decrease the pH of a solution. NH4+(aq) NH3(aq) + H+(aq) 4. With the exception of Group 1A metal ions and Ca2+, Sr2+, and Ba2+, metal ions will cause a decrease in the pH of a solution.

  16. Acid-Base Properties of Salt Solutions Will a solution of Na2HPO4 be acidic or basic? Two reactions are possible, one where HPO42- acts as an acid(Na+ is a spectator): HPO42-(aq) PO43-(aq) + H+(aq) and one where HPO42- acts as a base: HPO42-(aq) + H2O(l) H2PO4-(aq) + OH-(aq) HPO42- is a stronger base. The solution of Na2HPO4 will be basic.

  17. Factors That Affect Acid Strength 1. The polarity of the H-X bond. The more electronegative the X, the stronger the acid. H—X H atoms in nonpolar bonds (such as C-H bonds) are not acidic.

  18. Factors That Affect Acid Strength 2. The strength of the H-X bond. The weaker the bond, the stronger the acid. HF is a weak acid. But the F is the most electronegative element! The H-F bond is so strong that it predominates over the polarity. 3. The stability of the conjugate base. The more stable the conjugate base, the stronger the acid.

  19. Strength of Carboxylic Acids Carboxylic acids contain the carboxyl group -COOH. H :O: Ι║ acetic acid is H—C—C—Ö—H Ι H .. The C=O bond makes the O-H bond more polar. Furthermore, resonance stabilizes the RCOO- anion that is the conjugate base of the acid. Other electronegative atoms near the COOH group will further increase its acidity (CF3COOH, pKa = 0.2, is stronger than CH3COOH, pKa = 4.74).

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