Aim # 5: How do we determine the acidity (or basicity) of a solution? H.W. # 5 Study pp. 659-666 (up to sec. 14.5) Ans. ques. p. 703 # 51,54-56 How would you prepare 1600 mL of a pH=1.50 solution using concentrated (12M) HCl? Show your work. p. 710A (MC) # 5 Do Now: Zumdahl (8th ed.) p. 694 # 150

I The dissociation (autoionization) of water Water is amphoteric – a substance that can act as an acid or a base. 2H2O(ℓ) ↔ H3O+(aq) + OH-(aq) H‒O + H‒O ↔ H‒O‒H + + O‒H - | | | H H H More simply, H2O ↔ H+ + OH- and K = [H+][OH-] [H2O]

Because the molarity of pure water is constant, K[H2O] = [H+][OH-] = 1 Because the molarity of pure water is constant, K[H2O] = [H+][OH-] = 1.0 x 10-14 = Kw We define Kw as the ion-product constant (or dissociation constant) for water. At 250, Kw = [H+][OH-] = 1.0 x 10-14 Problem: The hydrogen ion concentration of a solution is [H+] = 3.5 x 10-4 M, at 250C. Is the solution acidic, neutral, or basic?

Ans: [H+][OH-] = 1.0 x 10-14 [OH-] = 1.0 x 10-14 = 1.0 x 10-14 [H+] 3.5 x 10-4 [OH-] = 2.9 x 10-11 M Because [H+] > [OH-], the solution is basic. Note: If [H+] > [OH-], a solution is acidic. If [H+] = [OH-], a solution is neutral. If [H+] < [OH-], a solution is basic. For pure water (or any neutral solution) [H+] = [OH-] (x)(x) = 1.0 x 10-14 x2 = 1.0 x 10-14 x = 1.0 x 10-7 = [H+] = [OH-]

II The pH scale 0 7 14 increasing neutral increasing acidity basicity

The pH of a solution indicates the concentration of hydronium ions in that solution. pH = -log [H+] or [H+] = 10-pH Problem: What is the pH of a solution whose [H+] Is 1.0 x 10-6? 3.5 x 10-10? Ans: pH = -log(1.0 x 10-6) = 0 – (-6.00) pH = 6.00 The solution is acidic.

Note: The number of decimal places in the log is equal to the number of significant figures in the original number. pH = -log(3.5 x 10-10) = -.54 – (-10.00) pH = 10.00 - .54 pH = 9.46 The solution is basic. Solution Type [H+] (M) [OH-] (M) pH Acidic >1.0 x 10-7 <1.0 x 10-7 <7.00 Neutral 1.0 x 10-7 1.0 x 10-7 7.00 Basic <1.0 x 10-7 >1.0 x 10-7 >7.00

Problem: What is the pH of a solution in which [OH-] = 4.3 x 10-11? Ans: [H+] = Kw = 1.0 x 10-14 = 2.3 x 10-4 M [OH-] 4.3 x 10-11 pH = -log(2.3 x 10-4) = -.36 –(-4.00) pH = 4.00 - .36 pH = 3.64

pOH = -log [OH-] and pKw = -logKw III pOH (and pKw) pOH = -log [OH-] and pKw = -logKw Since Kw = [H+][OH-] = 1.0 x 10-14 -log Kw = -log[H+]-log[OH-] = 14.00 pKw = pH + pOH = 14.00 for any aqueous solution at 250C How could we solve the last problem using pOH?

IV Strong Acids – ionize almost completely in aqueous solution. HA(aq) + H2O(ℓ) → H+(aq) + A-(aq) A. There are seven common strong acids: hydrobromic acid, HBr *sulfuric, H2SO4 hydrochloric acid, HCl chloric, HClO3 hydroiodic acid, HI perchloric, HClO4 nitric, HNO3 *diprotic – more than one acidic proton e.g. HNO3(aq) + H2O(ℓ) → H3O+(aq) + NO3-(aq) OR HNO3(aq) → H+(aq) + NO3-(aq) Equilibrium lies all the way to the right.

B. Calculating [H+] and pH for solutions of strong acids Problem: What is the pH of a 0.025 M solution of HNO3? Ans: Because ionization is complete [H+] = [NO3-] = 0.025 M pH = -log(0.025) = 1.60 Problem: An aqueous solution of HCl has a pH of 4.16. What is the concentration of the acid?

Ans: -log[H+] = 4.16 and log[H+] = -4.16 [H+] =10-4.16 = 6.9 x 10-5 M Practice Problems Zumdahl (8th ed.) p. 673 # 57,55,60,149