1. We use the dissociation of water equation to find [OH^-].
   K_w = [H₃O⁺][OH^-] = 1.0 X 10^-14
   Solve for [OH^-]
   [OH^-] = (1.0 X 10^-14)/ [H₃O⁺]
   Plug in the molarity of HI and solve for OH^-.
   [OH^-] = (1.0 X 10^-14)/ [2.4 X 10^-3] = 4.17 X 10^-12 M.
    
2. pH = -log[H₃O⁺]
   Plug the molarity of the HCl in and solve for pH.
   pH = -log[0.0035] = 2.46
    
3. pH = -log[H₃O⁺]
   Plug in the pH and solve for [H3O+]
   5.65 = -log[H₃O⁺]
   Move the negative sign to the pH. -5.65 = log[H₃O⁺]
   10^-5.65=10^log[H₃O+] = 2.24 X 10^-6 M
    
4. pH + pOH = 14
   Solve for pH.
   14 - pOH = pH
   14 - 4.74 = pH = 9.26
    
5. There are several ways to do this problem.
   Answer 1.
   pH + pOH = 14
   Solve for pOH.
   pOH = 14 - pH
   pOH = 14 - 1.5 = 12.5
   When the pOH is solved, solve for the concentration by using log.
   pOH = -log[OH^-]
   12.5 = -log[OH^-]
   -12.5 = log[OH^-]
   10^-12.5 = 10^log[OH-] = 3.16 X 10^-13 M.
   
   Answer 2.
   pH = -log[H⁺]
   Plug in the pH and solve for the molarity of H+ of pepsin. 
   1.5 = -log[H⁺]
   -1.5 = log[H⁺]
   10^-1.5 = 10^{log[H+] =} [H⁺]= 0.32
   Use the concentration of H⁺ to solve for the concentration of OH^-. 
   [H⁺][OH^-] = 1.0 X 10^-14
   Plug in the [H⁺] and solve for [OH^-].
   [OH^-] = (1.0 X 10^-14)/[H₃O⁺]
   [OH^-] = (1.0 X 10^-14)/(0.32) = 3.125 X 10^-14 M

6. 

From the ionization of acetic acid,

CH₃COOH = CH₃COO^- + H⁺
0.100             0.0042     0.0042we conclude that
[H⁺] = [CH₃COO^-]
      = 0.0042.
Thus, pH = -log0.0042 = 2.376.The equilibrium constant of ionzation,

        (0.0042)²
K = ------------- = 1.78x10^-5
          1.000

7. (a)
8. (c)
9. (a)
10. 

• The ions present are Na⁺ and OCl^- as shown by the following reaction:

NaOCl(s)?Na+(aq)+OCl?(aq)

While Na⁺ will not hydrolyze, OCl^- will (remember that it is the conjugate base of HOCl). It acts as a base, accepting a proton from water.

OCl?(aq)+ H2O(l)?HOCl(aq)+OH?(aq)

Na⁺ is excluded from this reaction since it is a spectator ion.

Therefore, with the production of OH^-, it will cause a basic solution, raising the pH above 7.

pH>7

• The KCN_(s) will dissociate into K⁺_(aq) and CN^__(aq) by the following reaction:

 KCN(s)?K+(aq) +CN?(aq)

K⁺ will not hydrolyze, but the CN^- anion will attract an H⁺away from the water:

CN?(aq)+H2O(l)?HCN(aq)+ OH?(aq)

Because this reaction produces OH^-, the resulting solution will be basic, causing a pH>7.

pH>7

• The NH₄NO_{3 (s)} will dissociate into NH₄⁺ and NO₃^- by the following reaction:

NH_4NO_3_{(s)} \rightarrow NH_4^+_{(aq)} + NO_3^-_{(aq)}

Now, NO₃^- won't attract an H⁺ because it is usually from a strong acid, meaning the K_b will be very small. However, NH₄⁺ will lose an electron, acting as an acid (NH₄⁺ is the conjugate acid of NH₃) by the following reaction:

NH_4^+_{(aq)} + H_2O_{(l)}\rightleftharpoons NH_3_{(aq)} + H_3O^+_{(aq)}

This reaction produces hydronium ion, making the solution acidic, lowering the pH below 7.  pH<7

16. K_b = 1.7 × 10^?9; pK_b = 8.77

17. pH = 12.08, % ionization = 5.4%

18. 8.18

19. acid B < acid C < acid A (strongest)

20. 

1. 
   
   1. K_a = 6.3 × 10^?11; pK_a = 10.20
   2. K_a = 7.9 × 10^?7; pK_a = 6.10
   3. K_a = 0.50; pK_a = 0.30
   4. K_a = 2.5 × 10^?13; pK_a = 12.60
   5. K_a = 3.2 × 10^?17; pK_a = 16.50

      21. K_a = 6.3 × 10^?10 pK_a = 9.20

22. 

1. neutral
2. acidic
3. basic (due to the reaction of HS^? with water to form H₂S and OH^?)