Unit 3: Chemical Systems and Equilibrium
8.4 Calculations Involving Acidic
Solutions
Strong Acids
▪
Strong Acids:
➢ionize
➢NO
(splits up into ions) almost 100% in water
EQUILIBRIUM
➢mostly
ions in solution
➢amount
of HCl present is negligible
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
Calculating the pH of a Strong Acid
Calculate the pH of a 0.050 M HCl solution
Since HCl is a strong acid, it dissociates completely
HCl(aq) + H2O (l) → H3O+(aq) + Cl-(aq)
[H3O+] = 0.050 M
pH = -log[H3O+]
= -log(0.050)
= 1.30
Measuring Strengths of WEAK Acids, Ka
•
•
•
Ionization of weak acids is not complete, as shown by the reversible
arrow. (Some unionized acid remains in solution).
CH3COOH(aq) +H2O(l) ↔ H3O+(aq) + CH3COO-(aq)
Since this is an equilibrium, the extent of ionization can be
represented using an equilibrium constant.
The equilibrium constant is given the subscript “a” to indicate that
the equilibrium involves an acid ionization; Ka is called an acid
ionization constant
Measuring Strengths of Weak Acids, Ka
CH3COOH(aq) +H2O(l) ↔ H3O+(aq) + CH3COO-(aq)
Weak acid
▪
Stronger acids have a large Ka
▪
Weaker acids have a small Ka
Examples of Ka Values for Some Weak Acids:
ACID
FORMULA
Ka
Nitrous Acid
HNO2
7.2 x 10-4
Acetic Acid
CH3COOH
1.8 x 10-5
Hydrocyanic Acid
HCN
6.2 x 10-10
Percent Ionization
• pH of weak acids tends to be closer to 7, whereas strong acids read a
much lower number.
• Most weak acids ionize at less than 50% (only 50% of it ionizes into
ions)
percentage ionization = concentration of acid ionized
initial concentration of acid
x 100%
Example 1
The pH of a 0.10mol/L methanoic acid solution is 2.38. Calculate the
percent ionization of methanoic acid.
• Step 1: Calculate the concentration of H+ using the pH.
• Step 2: Calculate percent ionization.
Example 2
Calculate the acid ionization constant Ka, of acetic acid if a 0.100 mol/L
solution at equilibrium at SATP has a percent ionization of 1.3%.
1. Write out the reaction:
2. Use an ICE table to organize data
Example 2 Con’d
Calculating pH for Weak Acid Solutions
For ionization of any acid, HA:
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
*NOTE: Use HUNDRED RULE to Avoid Quadratic Equations
If Ka is small, x will usually be small as well because not much ionization of the
acid will occur. If x is negligible relative to [HA]initial, the we can approximate:
([HA]initial – x) = ([HA]initial - x)
i.e. HUNDRED RULE: when the initial concentration of acid is greater than 100
times the Ka for that acid, we can remove x.
Example 3: Calculating pH for Weak Acid Solutions
Calculate the pH of 1.0 M solution of hydrofluoric acid in water.
▪ Step 1: Determine the relevant chemical equation(s) and find the
Ka values
▪ Step 2: Identify all species (Acid-Base conjugate pairs)
▪ Step 3: Summarize the changes in concentrations using an ICE
chart.
▪ Step 4: Use the equilibrium expression to calculate the value for x.
▪ Step 5: Apply the approximation when possible
▪ Step 6: Solve the equation for x.
▪ Step 7: Determine the pH.
Example 3: Calculating pH for Weak Acid Solutions
(cont’d)
Calculate the pH of 1.0 M solution of hydrofluoric acid in water.
Example 3: Calculating pH for Weak Acid Solutions
(cont’d)
Calculate the pH of 1.0 M solution of hydrofluoric acid in water.
Example 3: Calculating pH for Weak Acid Solutions
(cont’d)
Calculate the pH of 1.0 M solution of hydrofluoric acid in water.
Polyprotic Acids
•
Monoprotic Acids: have only one ionizable hydrogen atom
•
•
Polyprotic Acids: more than one ionizable hydrogen atom
•
•
•
•
Ex: H2SO4(aq), H3PO4(aq)
Both strong and weak acids can be polyprotic
Ionization of polyprotic acids occurs in multiple steps – one
hydrogen is ionized in each step
Each step has its own ionization constant (Ka) that is numbered
based on the step
•
•
Ex: HCl(aq), HNO3(aq)
Ex: Ka1, Ka2
Most often the initial acid is the strongest and subsequent
acids are weaker, as such we use the Ka for the first reaction to
calculate pH
Example 4: Calculating pH of a Polyprotic Acid
Calculate the pH of 0.10 M solution of ascorbic acid (vitamin C),
H2C6H6O6(aq). Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12
Example 4: Calculating pH of a Polyprotic Acid
Calculate the pH of 0.10 M solution of ascorbic acid (vitamin C),
H2C6H6O6(aq). Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12
Example 4: Calculating pH of a Polyprotic Acid
Calculate the pH of 0.10 M solution of ascorbic acid (vitamin C),
H2C6H6O6(aq). Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12
Additional Resources
• pH of Weak Acids and Bases - Percent Ionization - Ka & Kb – YouTube
• pH Calculations Involving Weak Acids – YouTube
• AP® Chemistry Weak Acid Equilibrium Questions - YouTube
Homework
• Read Section 8.4 – Pgs 512 - 524
• Pg 516 #1, 2
• Pg 520 #1, 2
• Pg 521# 1, 2
• Pg 524 #1
• Pg 525 #7