COMPARISON OF STRONG AND WEAK ACIDS (of the same concentration, eg mol/l) Characteristic Strong Acid Weak Acid. Equal to Solution Concentration

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1 1 COMPARISON OF STRONG AND WEAK ACIDS (of the same concentration, eg 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 K a High value Low value Electrical Conductivity High (strong electrolytes) Lower(weak electrolytes) Reaction with metals and carbonate compounds Fast Slower ph value Low (1 or less) 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) + H 2 O (l) H 3 O + (aq) + A - (aq) % Reaction = H 3 O + (aq) x 100 HA (aq) Example: 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? 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 CH 3 COOH has a ph of 2.67 vs mol/l of HCl has a ph of 1.0

2 2 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. K a = acid ionization constant (dissociation) K b = base ionization constant (dissociation) Examples: 1. For a weak acid such as acetic acid: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) The expression for the equilibrium constant is the ratio of ionized (dissociated) form to unionized (undissociated) form: K a = CH 3 COO - (aq) H 3 O + (aq) CH 3 COOH (aq) = 1.8 x 10-5 mol/l for a 0.10 mol/l solution of acetic acid at 25 o C (298K) 2. For a weak base such as ammonia: K b = NH 4 + (aq) OH - (aq) NH 3(aq) NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) = 1.7 x 10-5 mol/l for a 0.10 mol/l solution of ammonia at 25 o C (298K) Calculations involving K a and K b 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 K a or K b = mol/l 4. K a x K b = K w A. Finding K a or K b when equilibrium concentration of one species is given Example 1: Calculate the K a of a 0.75 mol/l solution of nitrous acid that has an equilibrium hydronium ion concentration of 0.36 mol/l.

3 3 Example 2: A 0.35M sodium hydrogen carbonate solution has an equilibrium concentration of carbonic acid of 0.11 mol/l. Calculate the K b for the base. B. Finding K a or K b when ph or poh is given Example 3: A 0.25 mol/l solution of carbonic acid was found to have a ph of What is the K a for the carbonic acid? Example 4: The poh of a mol/l solution of sodium propanoate, NaC 2 H 5 COO, is found to be Calculate K b for the propanoate ion.

4 4 C. Finding equilibrium concentrations when K a or K b 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. a) If the neglected quantity (x) is less than the retained quantity, then the 5% rule is valid. b) If the value of K a or K b is 1.00 x 10-7 or smaller the 5% rule always applies; c) If the value of K a or K b is 1.00 x 10-6 or larger check the 5% rule. Example 5: Calculate the equilibrium concentration of hydronium ions in a 0.56 M solution of HClO 2(aq) if the K a of the solution is 3.5 x 10-6 mol/l. Example 6: Formic acid is a moderately weak monoprotic acid with a K a of 1.8 x Find the hydronium concentration in a M solution of formic acid.

5 5 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 Example 8: Find the ph of a mol/l solution of hydrocyanic acid that has a K a of 2.7 x 10-4 mol/l.

6 6 Exercises: 1. Write the K a or K b expression for the following solutions: a) H 2 SO 3(aq) + H 2 O (l) b) Na 3 BO 3(aq) + HOH (l) 2. Suppose you measured the concentration of hydronium ions to be 1.6 x 10-4 mol/l in a 0.75 mol/l carbonic acid solution. Calculate the K a for the acid. 3. A 0.20 mol/l solution of an unknown acid, HA, has a hydronium ion concentration of 4.8 x 10-4 mol/l. a) Calculate the K a for this acid. b) Calculate the % reaction (%ionization) for this acid. 4. Find the K b for the benzoate ion in a 0.74 mol/l solution of NaC 6 H 5 COO (aq) whose poh is Calculate the K a for a 0.39 mol/l solution of HClO 3(aq) whose ph is Calculate the ph of a mol/l carbonic acid solution whose K a is 7.8 x 10-3 mol/l. 7. Find the poh and ph of a mol/l sodium glutamate solution, NaC 5 H 8 NO 4(aq), whose K b for the glutamate ion is 2.9 x 10-4 mol/l. 8. Calculate the percent reaction of a 1.00M solution of hydrocyanic acid, HCN, if its K a value is 6.2 x Calculate the hydroxide ion concentration and the poh of a mol/l ammonia solution whose K b = 1.77 x Calculate the ph of a solution whose initial hypochlorous acid concentration is 0.200M and whose K a value is 2.9 x What simplifying assumptions can be made in this problem?