1. No category

Sample Problem - Predicting the Direction of an Acid-Base Reaction
Predict the direction in which the following reaction will proceed. State whether reactants or products are
favoured at equilibrium, and briefly defend your reasoning.
SO4 2− (aq) + CH3COOH(aq) ↔ HSO4 − (aq) + CH3COO− (aq)
Solution: Identify the acid, base, conjugate acid and conjugate base.
SO4 2− (aq)
+
CH3COOH(aq)
↔
HSO4 − (aq)
+
CH3COO− (aq)
Base
acid
conjugate acid
conjugate base
Use Figure 14.12 to assess the relative strengths of the acids and bases. You can see that HSO4 − is above
CH3COOH, so HSO4 − is a stronger acid than CH3COOH. Similarly, CH3COO− is lower than SO4 2−, so
CH3COO− is a stronger base. The stronger acid and stronger base form the weaker acid and the weaker
base, so the direction proceeds in that direction—that is, to the left.
Please complete questions 10-11
Sample Problem - Determining [H3O+] and [OH–]
Find [H3O ] and [OH ] in each solution.
(a) 2.5 mol/L nitric acid
(b) 0.16 mol/L barium hydroxide
+
−
Solution: You know that nitric acid is a strong acid and barium hydroxide is a strong base. i) Since both
dissociate completely in aqueous solutions, you can use their molar concentrations to determine [H3O+] or
[OH−]. ii)You can then find the concentration of the other ion using Kw:
Kw = [H3O+] [OH−]
= 1.0 × 10−14
(a) [HNO3] = 2.5 mol/L, so [H3O+] = 2.5 mol/L
Kw = [H3O+] [OH−]
1.0 × 10−14 = 2.5 mol/L [OH−]
1.0 × 10−14 = [OH−]
2.5 mol/L
4.0 × 10−15 mol/L = [OH−]
(b) Ba(OH)2(aq) → Ba2+(aq) + 2OH−(aq)
Each mole of Ba(OH)2 in solution forms two moles of OH− ions.
∴ [OH−] = 2 mol OH × 0.16 mol/L = 0.32 mol/L
1 mol Ba(OH)2
Kw = [H3O+] [OH−]
1.0 × 10
= [H3O+] (0.32 mol/L)
1.0 × 10−14 = [H3O+]
0.32 mol/L
3.1 × 10−14 mol/L = [H3O+]
−14
Check Your Solution
For a solution of a strong acid, as in part (a), [H3O+] should be greater than 1.0 × 10−14 and [OH−] should be less
than 1.0 × 10−14. For a solution of strong base, [OH−] should be greater than, and [H3O+] should be less than, 1.0
× 10−14.
Sample Problems – Calculating the pH of a solution
Calculate the pH of a solution with [H3O+] = 3.8 × 10−3 mol/L.
pH = -log[H3O+] = -log (3.8 × 10 -3) = 2.420216403 = 2.42
(With pH and pOH, only consider what follows the decimal as significant. Therefore, the above answer includes
2 significant digits.)
Check Your Solution
[H3O+] is greater than 1.0 x 10−7 mol/L. Therefore, the pH should be less than 7. The solution is acidic.
Sample Problem – Calculating pH and pOH
A liquid shampoo has a hydroxide ion concentration of 6.8 10 x-5 mol/L at 25°C.
(a) Is the shampoo acidic, basic, or neutral?
Compare [OH−] in the shampoo with [OH−] in neutral water at 25°C. [OH−] in neutral water is 1 x 10 −7
mol/L. Since the [OH-] in shampoo is greater than this, the shampoo must be basic.
(b) Calculate the hydronium ion concentration.
Kw = [H3O+][OH-]
1 x 10-14 = [H3O+](6.8 x 10-5 mol/L)
1 x 10-14 = [H3O+](6.8 x 10-5 mol/L)
6.8 x 10-5
1.5 x 10-10 mol/L = [H3O+]
(c) What is the pH and the pOH of the shampoo?
pH = -log[H3O+] = - log (1.5 x 10-10) = 9.83
pOH = -log[OH-] = - 10 log (6.8 x 10-5) = 4.17
Check Your Solution
pH + pOH = 14
(4.17 + 9.83 = 14)
Complete questions (#16-19)
Sample Problem – Determining pOH, [H3O+], and [OH-]
If the pH of urine is outside the normal range of values, this can indicate medical problems. Suppose that the pH
+
of a urine sample was measured to be 5.53 at 25°C. Calculate pOH, [H3O ], and [OH ] for the sample.
Solution
pH given = 5.53
pOH = 14.00 – 5.53 = 8.47
+
-pH
[H3O ] = 10
-
[OH ] = 10
-pOH
-5.53
= 10
= 10
-8.47
-6
-6
= 2.95121 x 10 M = 3.0 x 10 M
-9
-9
= 3.3884 x 10 M= 3.4 x 10 M
Check Your Solution
+
-6
-9
-14
Kw= [H3O ][OH ] = (3.0 x 10 M)(3.4 x 10 M) = 1.0 x 10
This value equals the expected value for Kw at 25°C.