1. Chapter Fifteen Acids and Bases

2. Chapter Fifteen/ Acids and Bases acids is a substances that ionize in water to produce H + ions HCl (aq) → H + (aq) + Cl ‐ (aq) bases is a substances that ionize in water to produce OH - ions. NaOH (aq) → Na + (aq) + OH ‐ (aq) An acid neutralizes a base H + (aq) + OH ‐ (aq) → H 2 O(ℓ) The Acid-Base Properties of Water

3. Chapter Fifteen/ Acids and Bases Water has the ability to act either as an acid or as a base. Water undergo ionization to a small extent this reaction is sometimes called the autoionization of water. The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 °C, K w = 1.0  10 −14 The Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH -

4. Chapter Fifteen/ Acids and Bases For water: Because water is neutral then The Acid-Base Properties of Water [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic

5. Chapter Fifteen/ Acids and Bases Example: Calculate the [H + ] ions in aqueous ammonia, [OH - ] =0.0025 M? THUS [H + ] < [OH - ] therefore the solution is basic The Acid-Base Properties of Water

6. Chapter Fifteen/ Acids and Bases Because the concentrations of H + and OH - ions in aqueous solutions are frequently very small numbers and therefore inconvenient to work with, Soren Sorensen in 1909 proposed a more practical measure called pH. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L). For [OH] pH—A Measure of Acidity

7. Chapter Fifteen/ Acids and Bases pH—A Measure of Acidity AcidicBasic Increase the acidityIncrease the basisty pH is inversely proportional to [H + ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C

8. Chapter Fifteen/ Acids and Bases Example The concentration of H + ions in a bottle of vinegar was 3.2 x 10 -4 M right after the cork was removed. Only half of the vinegar was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 x 10 -3 M. Calculate the pH of the vinegar on these two occasions. pH—A Measure of Acidity

9. Chapter Fifteen/ Acids and Bases Example The pH of rainwater collected in a certain region of the northeastern Saudi Arabia on a particular day was 4.82. Calculate the H + ion concentration of the rainwater. pH—A Measure of Acidity

10. Chapter Fifteen/ Acids and Bases Example In a NaOH solution [OH - ] is 2.9 x 10 -4 M. Calculate the pH of the solution? pH—A Measure of Acidity

11. Chapter Fifteen/ Acids and Bases Strong acid (or base) have 100 % dissociation. Weak acid (or base) have incomplete dissociation. Weak Acids and Acid Ionization Constants HCl (s) H + (aq) + Cl - (aq) H2OH2O CH 3 COOH CH 3 COO - (aq) + H + (aq)

12. Chapter Fifteen/ Acids and Bases The acid ionization constant (K a ), is the equilibrium constant for the ionization of an acid. At a given temperature, the strength of the acid HA is measured quantitatively by the magnitude of K a. The larger K a, the stronger the acid—that is, the greater the concentration of H + ions at equilibrium due to its ionization Weak Acids and Acid Ionization Constants

13. Chapter Fifteen/ Acids and Bases Example What is the pH of a 0.5 M HF solution (at 25 0 C) if K a = 7.1x10 -4 ? Weak Acids and Acid Ionization Constants HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] HF H + F - Initial (M) 0.5 0 0 Change -x x x Equilibrium 0.5 - x x x K a = x2x2 0.50 - x = 7.1 x 10 -4 0.50 – x  0.50 K a << 1

14. Chapter Fifteen/ Acids and Bases When can I use the approximation? When x is less than 5% of the value from which it is subtracted. Weak Acids and Acid Ionization Constants K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = 0.019 M pH = -log [H + ] = 1.72 x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok.

15. Chapter Fifteen/ Acids and Bases Example What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? Weak Acids and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] HA H + A - Initial (M) 0.122 0 0 Change -x x x Equilibrium 0.122 - x x x K a = x2x2 0.122 - x = 5.7 x 10 -4 0.122 – x  0.122 K a << 1

16. Chapter Fifteen/ Acids and Bases When can I use the approximation? When x is less than 5% of the value from which it is subtracted. Weak Acids and Acid Ionization Constants Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok. K a = x2x2 0.122 - x = 5.7 x 10 -4 X 2 + 0.00057X – (6.95 x 10 -5 )= 0 ax 2 + bx + c =0 X 2 = (6.95 x 10 -5 ) - 0.00057X

17. Chapter Fifteen/ Acids and Bases Weak Acids and Acid Ionization Constants x = 0.0081x = - 0.0081 HA H + A - Initial (M) 0.122 0 0 Change -x x x Equilibrium 0.122 - x x x [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09

18. Chapter Fifteen/ Acids and Bases Example The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the K a of the acid? Weak Acids and Acid Ionization Constants HCOOH (aq) H + (aq) + HCOO - (aq) K a = [H + ][HCOO - ] [HCOOH] HCOOH H + HCOO - Initial (M) 0.1 0 0 Change -x x x Equilibrium 0.1 - x x x

19. Chapter Fifteen/ Acids and Bases Weak Acids and Acid Ionization Constants HCOOH H + HCOO - Initial (M) 0.1 0 0 Change -4.1X10 -3 4.1X10 -3 4.1X10 -3 Equilibrium 0.1 - 4.1X10 -3 4.1X10 -3 4.1X10 -3 K a = [H + ][HCOO - ] [HCOOH]

20. Chapter Fifteen/ Acids and Bases K a indicates the strength of an acid. Another measure of the strength of an acid is its percent ionization. The stronger the acid, the greater the percent ionization. Weak Acids and Acid Ionization Constants percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration

21. Chapter Fifteen/ Acids and Bases Example Calculate the percent ionization of hydrofluoric acid at the concentrations of 0.50 M if K a = 7.1x10 -4 ? Weak Acids and Acid Ionization Constants HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] HF H + F - Initial (M) 0.5 0 0 Change -x x x Equilibrium 0.5 - x x x K a = x2x2 0.50 - x = 7.1 x 10 -4 0.50 – x  0.50 K a << 1

22. Chapter Fifteen/ Acids and Bases Weak Acids and Acid Ionization Constants K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 x 2 = 3.55 x 10 -4 x = 0.019 M x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. [H + ] = 0.019 M percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% 0.19 0.5 x 100% percent ionization == 3.8%

23. Chapter Fifteen/ Acids and Bases Example A 0.040 M solution of a monoprotic acid is 14 percent ionized. Calculate the ionization constant of the acid.? Weak Acids and Acid Ionization Constants percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% HA H + A - Initial (M) 0.04 0 0 Change -0.0056 0.0056 0.0056 Equilibrium 0.04 – 0.0056 0.0056 0.0056

24. Chapter Fifteen/ Acids and Bases Weak Acids and Acid Ionization Constants K a = [H + ][A - ] [HA]

25. Chapter Fifteen/ Acids and Bases The ionization of weak bases is treated in the same way as the ionization of weak acids. The base ionization constant (K b ), is the equilibrium constant for the ionization of a base. At a given temperature, the strength of the base BA is measured quantitatively by the magnitude of K b. The larger K b, the stronger the base—that is, the greater the concentration of OH - ions at equilibrium due to its ionization In solving problems involving weak bases, we follow the same procedure we used for weak acids. The main difference is that we calculate [OH - ] first, rather than [H + ]. Weak Bases

26. Chapter Fifteen/ Acids and Bases Example What is the pH of a 0.40 M ammonia solution if K b = 1.8x10 -5 ? Weak Bases NH 3 NH 4 + OH - Initial (M) 0.4 0 0 Change -x x x Equilibrium 0.4 - x x x K b = x2x2 0.4 - x = 5.7 x 10 -4 0.4 – x  0.4 K b << 1

27. Chapter Fifteen/ Acids and Bases When can I use the approximation? When x is less than 5% of the value from which it is subtracted. Thus Weak Bases Ka  Ka  x2x2 0.40 = 1.8 x 10 -5 x 2 = 7.2 x 10 -6 x = 2.7 x 10 -3 M [OH - ] = 2.7 x 10 -3 M pOH = -log [OH - ] = 2.57 x = 2.7 x 10 -3 M 2.7 x 10 -3 M 0.40 M x 100% = 0.68% Less than 5% Approximation ok.