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Calculations Involving Acids
Section 8.
Homework Pg. 516 # Pg. 520 #1, 2 Pg. 521 #1, 2 Pg. 524 #1, 2 Pg. 525 #1-10a
Types of Calculations
• K a
• pH
• % ionization
…or some combination of all of the above!
Solutions of Strong Acids
Complete ionization occurs:
HA H+^ + A-
(before ionization) 1.0 mol
(after ionization) 1.0 mol
(after ionization) 1.0 mol
100%
Therefore, [HA] = [H +] = [A-^ ]
Example 1. Find the [H +], [OH -^ ], pH and pOH of
a 0.042 mol/L HNO3 (aq) solution.
This reaction proceeds according to the equation:
HNO 3 (aq) H+ (aq) + NO 3 - (aq)
Before getting started, consider the MAJOR entities in solution:
There are actually TWO processes that can contribute to [H +] : (1) ionization of HNO 3 Ka = infinitely large (2) autoionization of water: H 2 O H+^ + OH-^ Kw = 1.0 x 10-
H +^ NO 3 -^ H 2 O
Ignore process (2) because Ka of HNO 3 is so much larger ∴ HNO 3 is a much stronger acid; contribution of H 2 O is negligible
Example 1. Find the [H +], [OH -^ ],
pH and pOH of a 0.042 mol/L
HNO3 (aq) solution.
Strategy for strong acid calculations:
- Assume that [H +^ ] = [HA]
- Perform necessary calculations using [H +^ ]
Example 1. Find the [H +], [OH -^ ],
pH and pOH of a 0.042 mol/L
HNO3 (aq) solution.
Strategy for strong acid calculations:
- Assume that [H +^ ] = [HA]
- Perform necessary calculations using [H +^ ]
Find [H+] Since HNO 3 is a strong acid, [H+] = [HNO 3 ] = 0.042 M
Find pH pH = - log[H+] = - log(0.042) pH = 1.
Find pOH pH + pOH = 14. pOH = 14.00 – pH pOH = 14.00 – 1. pOH = 12.
Find [OH-] [OH-] = 10-pOH = 10 -12. [OH-] = 2.4 × 10 -13^ mol/L
Practice.
Pg. 513 #1, 2
1. Find [OH-^ ] of a 0.0700 mol/L HCl solution. Ans: 1.43 x 10-13^ M
2. A 2.00-L HBr solution contains 0.070 mol of acid.
Find pH and pOH. Ans: pH = 1.46, pOH = 12.
Solutions of Weak Acids
Weak acids only partially ionize in water.
- Percent ionization describes how much of the original acid ionizes to produce H +.
For the general weak acid ionization reaction
% ionization =
[H+]
[HA] 0
×100%
HA H+^ + A
where [H +^ ] is the concentration of ionized acid, and [HA] 0 is the initial acid concentration.
Example 2. Finding % Ionization Using pH
Calculate the percent ionization of propanoic acid, HC 3 H 5 O 2 ,
if a 0.050 mol/L solution has a pH of 2.78.
HC 3 H 5 O 2 ⇌ H +^ + C 3 H 5 O 2 -
% ionization =
[H+^ ]
[HC 3 H 5 O 2 ] 0 ×100%
= 1.7×^10
− 3 0.050 ×100% % ionization = 3.3%
Calculate % ionization
^ Use pH to find [H+]
Write the ionization equation
[H +] = 10-2.
[H+] = 1.7 x 10 -3^ mol/L
the quantity of H+ from the ionization of HC 3 H 5 O2.
Example 3. Finding Ka using % Ionization
Calculate the K a of hydrofluoric acid, HF, if a 0.100 mol/L solution
at equilibrium has a percent ionization of 7.8%.
Ka =
H +^ [F − ]
HF
Requires all concentrations at equilibrium
HF H+^ F -
I 0.100 0 0
C - x + x + x E 0.100 – x x x
HF
Write the ionization equation H+^ + F-
^ Write Ka expression
Set up ICE table
HF H+^ F -
I 0.100 0 0
E 0.100 – x x x
Given: % ionization = 7.8%
RTF: Ka
x = 0.078 × 0.100 mol/L x = 0.0078 mol/L = [H+] = [F-]
[HF] = 0.100 – x = 0.0922 mol/L
Ka =
H+^ [F − ]
HF =
Ka = 6.6 × 10 -
0.0922 mol/L 0.0078 mol/L (^) 0.0078 mol/L
Plug values into K (^) a
Use % ionization to find [H+] and [F - ]
^ Use^ x^ to find [HF]
Homework
Pg. 516
Pg. 520 #1, 2
Pg. 521 #1, 2
Pg. 524 #1, 2
Pg. 525 #1-10a
