1. Aim # 5: How do we determine the acidity (or basicity) of a solution?
H.W. # Study pp (up to sec. 14.5) Ans. ques. p. 703 # 51, 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

2. I The dissociation (autoionization) of water
Water is amphoteric – a substance that can act as an acid or a base H2O(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]

3. 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 = 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?

4. Ans: [H+][OH-] = 1.0 x 10-14
[OH-] = 1.0 x = 1.0 x [H+] x 10-4
[OH-] = 2.9 x 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-]

5. II The pH scale
increasing neutral increasing acidity basicity

6. 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 = The solution is acidic.

7. 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 = pH = The solution is basic.
Solution Type [H+] (M) [OH-] (M) pH Acidic >1.0 x <1.0 x < Neutral x x Basic <1.0 x >1.0 x >7.00

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

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

10. 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, H2SO hydrochloric acid, HCl chloric, HClO hydroiodic acid, HI perchloric, HClO nitric, HNO *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.

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

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