1. Calculations Involving Acids and Bases Section 18.1

2. Pure Water In aqueous solutions, water is in equilibrium: H 2 O H + (aq) + OH - (aq) ΔH = +57 kJ mol -1 Water at 25º C has a concentration of H + and OH - that are both equal to 1.00 x 10 -7 mol dm -3 At 25º, the product of the concentrations is always 1.00 x 10 -14 mol 2 dm -6 K w is known as the dissociation constant (or ionic product constant) of water: K w = [H + ] [OH - ] = 1.00 x 10 -14 mol 2 dm -6 (at 25º)

3. Equilibrium Forward reaction is endothermic (involves breaking bonds) As temperature is raised, equilibrium shifts to the right and the equilibrium constant increases At higher temps, [H + ] >10 -7 mol dm -3, pH <7, even though it is still neutral [H + ] [OH - ] ex. at 50º C, [H + ] = 3.05 x 10 -7 mol dm -3 = [OH - ] and the pH of the pure water is 6.5

4. Calculating Concentration of an Ion If you know the concentration of the hydrogen ions, you can divide 1.00 x 10 -14 by that concentration to get the concentration of the hydroxide ions. ex. What is the hydroxide ion concentration if the hydrogen ion concentration is 3.4 x 10 -3 ? 1.00 x 10 -14 = 2.94 x 10 -12 mol dm -3 3.4 x 10 -3

5. More Math Fun! The pH of a solution depends on the concentration of hydrogen ions pH = -log [H + ] and [H + ] = 10 -pH What is the pH of a solution that has a [H + ] of 4.6 x 10 -5 mol dm -3 ? pH = -log(4.6 x 10 -5 ) pH = 4.34 pH is reported to 2 decimal places

6. Even More! What is the pH of a 0.00100 mol dm -3 NaOH solution? [OH - ] = 1.00 x 10 -3 mol dm -3 [H + ] = 1.00 x 10 -14 = 1.00 x 10 -11 mol dm -3 1.00 x 10 -3 pH = -log 1.00 x 10 -11 mol dm -3 pH = 11 (obviously basic)

7. pOH Defined and calculated by: pOH = -log [OH - ] and [OH - ] = 10 -pOH What is the pOH of a solution with an [OH - ] of 5.98 x 10 -9 mol dm -3 ? pOH = -log (5.98 x 10 -9) = 8.22 pk w = -log K w pH + pOH = pK w = 14.00 (at 25º C)

8. KaKa Consider a weak acid in equilibrium: CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) The equilibrium constant for this reaction is known as the acid dissociation constant, K a K a = [H + ] [CH 3 COO - ] mol dm -3 [ CH 3 COOH] The greater the K a, the stronger the acid

9. pK a pK a = -log K a and K a = 10 -pKa K a of ethanoic acid is 1.74 x 10 -5 mol dm -3 at 298K, so its pK a is 4.76 pK a relates to the strength of the acid, greater the pK a value, the weaker the acid Varies with temperature

10. Calculating K a Consider HA (aq) H + (aq) + A - (aq) Initial conc. a 0 0 Equil. Conc. a – x x x K a = [H + ] [A - ] = x · x = x 2 [HA] a – x a – x If the acid is weak then x<< a, so it is almost equal to a, then K a ≈ x 2 (at 25º C) a

11. Assumption The previous information is valid if the pH is < 6 so that [H + ] = [A - ] (hydrogen ions from water molecules is neglected)

12. Example A 0.0100 mol dm -3 solution of a weak acid has a pH of 5.00. What is the dissociation constant of the acid? (K a ) If pH is 5.00, then [H + ] = [A - ] = 1.00 x 10 -5 mol dm -3 K a ≈ x 2 /a = (1.00 x 10 -5 ) 2 0.0100 K a = 1.00 x 10 -8 mol dm -3

13. Calculating pH Benzoic acid has a pK a of 4.20. What is the pH of a 0.100 mol dm -3 solution of this acid? If pK a = 4.20, the K a = 6.31 x 10 -5 mol dm -3 K a ≈ x 2 /a then x 2 = K a ∙ a x 2 = 6.31 x 10 -5 ∙ 0.100 x = √ 6.31 x 10 -6 = 2.51 x 10 -3 mol dm -3 pH = -log (2.51 x 10 -3 ) = 2.60

14. Calculating Concentration What concentration of HF is required to give a solution of pH 2.00 and what percentatge of the HF is dissociated at this pH, if the dissociation constant of the acid is 6.76 x 10 -4 mol dm -3 ? Use 10 -pH to get [H + ] = 1.00 x 10 -2 Rearrange K a = x 2 /a to solve for a Initial concentration = 0.148 mol dm -3 % dissociation = 100 x 1.00 x 10 -2 / 0.148 = 6.76%

15. Validity of Previous Problem “x” is not much smaller than “a” Can use the original formula of x 2 /(a-x) to get a more accurate value of 0.158 mol dm -3

16. Weak Base Dissolved in Water B (aq) + H 2 O BH + (aq) + OH - (aq) K b = [BH + ] [OH - ] (base dissociation constant) [B] Treating it with similar assumptions to those for weak acids: K b ≈ y 2 where pK b = -log K b b

17. Calculation with a Weak Base What is the pH of a 0.0500 mol dm -3 solution of ethylamine (pK b = 3.40)? Find K b using 10 -pKb Use K b = y 2 /b to find y y = √ 1.99 x 10 -5 [OH - ] = y = 4.46 x 10 -3 pOH = -log 4.46 x 10 -3 = 2.40 pH = 14 – 2.40 = 11.60

18. Conjugate Acid-Base Pair K a x K b = K w (10 -14 ) at 25 degrees C pk a + pK b = 14.00 at 25 degrees C The stronger the acid (the greater the K a ) the weaker the base (the smaller the K b )