Chemistry

Name: _____________________________

COMPARISON OF STRONG AND WEAK ACIDS (of the same concentration, eg. 0.10 mol/L) Characteristic

Strong Acid

Weak Acid

% reaction with water (%dissociation/ionization)

100%

Less than 50% for most

Hydronium ion concentration

Equal to Solution Concentration

Less than Solution Concentration

Ka

High value

Low value

Electrical Conductivity

High (strong electrolytes)

Lower(weak electrolytes)

Reaction with metals and carbonate compounds

Fast

Slower

pH value

Low (< 2)

Higher

Indicator Color

Intense (deep shade)

Less Intense (lighter shade)

Note: The % reaction with water of a weak acid is the fraction of acid molecules that dissociate (ionize) compared with the initial concentration of the acid. Ie. For an acid of the general formula, HA HA(aq)

+

H2O(l)

% Reaction =

Example:



H3O+(aq) + A-(aq)

H3O+(aq) HA(aq)

x 100

The concentration of hydronium ions was found to be 3.3 x 10 -4 mol/L in a 0.25 mol/L carbonic acid solution. What is the % reaction of this acid?

1

Chemistry

Name: _____________________________

Strength vs. Concentration: A dilute solution of a strong acid will often have a greater hydronium ion concentration & lower pH than a concentrated solution of a weak acid. Eg:

2.0 mol/L CH3COOH has a pH of 2.67 vs. 0.10 mol/L of HCl has a pH of 1.0

ACID & BASE IONIZATION (DISSOCIATION) CONSTANTS These equilibrium constants are used to indicate the degree of ionization or dissociation and therefore strength of weak acids and weak bases ie. their reaction with water to produce hydronium ions or hydroxide ions, respectively. Ka = acid ionization constant (dissociation)

Kb = base ionization constant (dissociation)

Examples: 1. For a weak acid such as acetic acid: CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) The expression for the equilibrium constant is the ratio of ionized (dissociated) form to unionized (undissociated) form: Ka = CH3COO-(aq)  H3O+(aq) CH3COOH(aq)  = 1.8 x 10-5 mol/L for a 0.10 mol/L solution of acetic acid at 25oC (298K) 2.

For a weak base such as ammonia: NH3(aq) + H2O(l)  NH4+ (aq) + OH-(aq) Kb = NH4+ (aq) OH-(aq) NH3(aq) = 1.7 x 10-5 mol/L for a 0.10 mol/L solution of ammonia at 25oC (298K)

Calculations involving Ka and Kb NOTE: 1. Do not include water in expressions for equilibrium constants, only solutions (or gases). 2.

Do not include spectator ions in the reaction equation, but always include water as one of the reactants.

3.

The units for Ka or Kb = mol/L

4.

Ka x Kb = Kw

2

Chemistry

A.

Name: _____________________________

Finding Ka or Kb when equilibrium concentration of one species is given

Example 1: Calculate the Ka of a 0.75 mol/L solution of nitrous acid that has an equilibrium hydronium ion concentration of 0.36 mol/L.

Example 2: A 0.35M sodium hydrogen carbonate solution has an equilibrium concentration of carbonic acid of 0.11 mol/L. Calculate the Kb for the base.

B.

Finding Ka or Kb when pH or pOH is given

Example 3: A 0.25 mol/L solution of carbonic acid was found to have a pH of 3.48. What is the K a for the carbonic acid?

3

Chemistry

Name: _____________________________

Example 4: The pOH of a 0.157 mol/L solution of sodium propanoate, NaC 2H5COO, is found to be 4.96. Calculate Kb for the propanoate ion.

C.

Finding equilibrium concentrations when Ka or Kb is given:

The 5% Rule  Calculations involving equilibrium are not accurate and a 5% error is acceptable.  The quadratic equation must be considered, but can be avoided if the 5% rule is valid ie. if the percent reaction of the weak acid or base with water is ≤ 5%. 

If the percent reaction is > 5% the quadratic equation must be used to determined the unknown concentration (x)

Example 5: Calculate the equilibrium concentration of hydronium ions in a 0.56 M solution of HClO 2(aq) if the Ka of the solution is 3.5 x 10-6 mol/L.

4

Chemistry

Name: _____________________________

Example 6: Formic acid is a moderately weak monoprotic acid with a Ka of 1.8 x 10-4. Find the hydronium concentration in a 0.0010 M solution of formic acid.

D.

Finding pH or pOH:

Example 7: Find the pOH of a 0.10M solution of sodium benzoate that has a K b of 1.6 x 10-10.

Example 8: Find the pH of a 0.015 mol/L solution of hydrocyanic acid that has a K a of 2.7 x 10-4 mol/L.

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