4. Predicting the direction of a rxn
Predict the direction of the following reaction:
HNO2 + CN- D HCN + NO2-
Solution: You can either compare between the acids or the bases in both sides
Ka (HNO2) = 4.5*10-4 and Ka(HCN) = 4.9*10-10
HNO2 is a stronger acid which means it is a better proton donor and thus rxn will proceed to right.
Also can say CN- is a stronger base than NO2- , therefore CN- is a better proton acceptor and rxn will proceed to right.

5. Predict whether the equilibrium constant for the following reaction is greater / or less than 1.
CH3COOH + HCOO- D CH3COO- + HCOOH
This is the same as asking whether the reaction proceeds in the forward or backward direction.
Ka(CH3COOH) = 1.8*10-5 and Ka(HCOOH) = 1.7*10-4.
Formic acid is a better acid than acetic acid, therefore the rxn will proceed to left and K<1.

6. Predict whether the equilibrium constant for the following reaction is greater / or less than 1.
CH3COOH(aq) + H2O(l) D
H3O+(aq) + CH3COO−(aq)
H3O+(aq) is the strongest acid, it is stronger than acetic acid, therefore, rxn proceeds to left. Ka<1
Or can say: Acetate is a stronger base than H2O and thus it is a better proton acceptor, so the equilibrium favors the left side (K<1).

7. What does Ka measure?
Ka is a constant used by chemists to compare the relative strength of weak acids.
Many chemists use pKa because it is more convenient.
pKa = -log Ka
For example: the pKa of H2O is and the pKa of acetic acid is Which is the stronger acid?
The lower the pKa, the stronger the acid.

8. Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
HA (aq) H+ (aq) + A- (aq)
[H+][A-]
Ka =
[HA]
Ka is the acid ionization constant
weak acid
strength Ka

10. What is the pH of a 0.5 M HF solution (at 250C)?
Ka =
[H+][F-]
[HF]
= 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
HF (aq) H+ (aq) + F- (aq)
Initial (M)
0.50
0.00
0.00
Change (M) -x +x +x
Equilibrium (M) x x x
Ka = x2 x
= 7.1 x 10-4
Ka << 1
0.50 – x  0.50
Ka  x2
0.50
= 7.1 x 10-4
x2 = 3.55 x 10-4
x = M
[H+] = [F-] = M
pH = -log [H+] = 1.72

11. When can I use the approximation?
Ka << 1
0.50 – x  0.50
When x is less than 5% of the value from which it is subtracted.
x = 0.019
Less than 5%??
Approximation ok.
RE = (0.019/0.50)*100% = 3.8%

12. What is the pH of a 0.05 M HF solution (at 250C)?
Ka  x2
0.05
= 7.1 x 10-4
x = M
More than 5% !!!!
Approximation not ok.
0.006 M
0.05 M
x 100% = 12%
RE =
Must solve for x exactly using quadratic equation or method of successive approximation.

13. HClO(aq) + H2O(l) D H3O+(aq) + ClO-(aq)
Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H3O+] of a M HClO solution? Ka = 3.5 x 10-8
Solution:
HClO(aq) + H2O(l) D H3O+(aq) + ClO-(aq)
[H3O+] [ClO-]
Ka = = 3.5 x 10-8
[HClO]

14. [H3O+] = x = 6.61*10-5 M RE = {6.61*10-5/0.125) *100% = 0.053%
Concentration (M) HClO H3O ClO-
Initial
Change x x x
Equilibrium x x x
(x)(x)
Ka = = 3.5 x 10-8
assume x = 0.125
0.125-x
x2 = x x = 6.61 x 10-5
RE = {6.61*10-5/0.125) *100% = 0.053%
[H3O+] = x = 6.61*10-5 M

15. What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M)
0.122
0.00
0.00
Change (M) -x +x +x
Equilibrium (M) x x x
Ka = x2 x
= 5.7 x 10-4
Ka << 1
0.122 – x  0.122
Ka  x2
0.122
= 5.7 x 10-4
x2 = 6.95 x 10-5
x = M
More than 5%??
Approximation not ok. M
0.122 M
x 100% = 6.8%
RE =

16. [H+] = x = 0.0081 M pH = -log[H+] = 2.09 Ka = x2 0.122 - x
x x – 6.95 x 10-5 = 0
ax2 + bx + c =0
-b ± b2 – 4ac  2a
x =
x =
x =
[H+] = x = M
pH = -log[H+] = 2.09

17. For a monoprotic acid HA
Ionized acid concentration at equilibrium
Initial concentration of acid
x 100%
percent ionization =
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration

18. % Dissociated (also called % Ionized) Weak Acids
% ionization – a useful way of expressing the strength of an acid or base.
100% ionized a strong acid.
Only partial ionization occurs with weak acids.

19. Calculate the percent ionization of 0.10 M HCOOH. Ka = 1.7*10-4
Ka =x2/(0.1 – x), assume 0.1>>x
X= 4.1*10-3, RE ={4.1*10-3/0.10} * 100 = 4.1%
% ionization = (4.1*10-3/0.10) *100 = 4.1%
HCOO- H+ D
HCOOH
0.10
Initial X+ -x
Change x
0.1-x
Equil

20. Calculate the Percent dissociation of a 0.0100M Hydrocyanic
acid solution, Ka = 6.20 x
HCN(aq) + H2O(l) H3O+(aq) + CN- (aq)
HCN H3O CN-
Initial M Ka =
Change x x x
Final –x x x
Ka = = 6.20 x 10-10
Assume x =
Ka = = 6.2 x 10-10
x = x 10-6
% dissociation = x 100% = 0.025%
[H3O+][CN-]
[HCN]
(x)(x)
( x) x2
0.0100
2.49 x 10-6

21. Finding the Ka of a Weak Acid from the pH of its Solution
Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid.
Calculating [H3O+] :
[H3O+] = 10-pH = = 6.46 x 10-5 M
Concentration(M) HClO(aq) H2O(l) H3O+(aq) + ClO -(aq)
Initial
Change x x x
Equilibrium x x x
Assumptions: [H3O+] = [H3O+]HClO

22. since HClO is a weak acid, we assume 0.12 M - x = 0.12 M
HClO(aq) + H2O(l) H3O+(aq) + ClO -(aq)
x = [H3O+] = [ClO-] = 6.46 x 10-5 M
[H3O+] [ClO-]
(6.46 x 10-5 M)
(6.46 x 10-5 M)
Ka = = = 3.48 x 10-8
[HClO]
0.12 M
Checking:
RE = x 100 = %
6.46 x 10-5 M
0.12 M