CALCULATIONS INVOLVING STRONG BASIC SOLUTIONS
 When you are trying to understand the calculations involving strong basic solutions, we must
understand that strong bases dissociate
in water. It is also important to
note that the metal cation produced has no influence over the pH of the solution.
 As a review, what are two ways in which we can calculate pOH?
Example Problem 1: A chemist prepares a solution by dissolving 0.042 mol or strontium hydroxide
in 2.00 L water, Sr(OH)2(aq). Calculate the [H+], [OH-], pH and pOH
CALCULATIONS INVOLVING WEAK BASIC SOLUTIONS
When you are trying to understand the calculations involving weak acid solutions, you have to
remember that dissociation is often incredibly small (at most, 50%) and there is often a large
concentration of base not reacted and a small concentration of conjugate acid and cation. This is
just like any equilibrium reaction and may require the use of an ICE table. You will need Kb and if
not given to you, can be calculated from Ka of the conjugate acid. The Kb can be used to
determine the pH of a solution.
Example Problem 2: Calculate the pH of 4.5 mol/L solution of hydrazine. The Kb (on Appendix B5 –
Table 3) is 1.7 x 10-6
Given: [N2H4 (aq)] = 4.5 mol/L, Kb = 1.7 x 10-6
Write a reaction of the weak base in water:
Write a Kb expression for hydrazine:
N2H4(aq) + H2O(aq) ⇌ N2H5+(aq) + OH-(aq)
Set-Up an ICE Table for the reaction of hydrazine. You can write out the simplified reaction
without water.
I
C
E
N2H4(aq)
⇌
N2H5+(aq)
+
OH-(aq)
Check to see if we can simplify our expression with the
hundred rule:
Can we simplify the change in [N2H4], so that
4.5 – x ≈ 4.5?
You may be given a problem in which you have to determine the Kb first. In such a case, you can
use the Ka of the conjugate acid in the reaction. Since KaKb = 1.0 x 10-14 (Kw), you can solve for Kb
and then proceed with calculating the pH. An example of this more extensive problem is on Page
527 – Tutorial 2.
HOMEWORK: Read Pages 526-529 and do questions #1ac, 2 bc, 3a, 4a, 5, 7