Lesson 8.5: Calculations Involving Basic Solutions: Strong and Weak Bases

Acid–Base Equilibrium

Calculations Involving Basic Solutions

This section explores the calculation methods for determining the properties of solutions containing strong and weak bases. The approach parallels that used for acids, with appropriate substitutions for base-related quantities such as Kb, [OH–], and pOH.

Calculations Involving Solutions of Strong Bases

• Strong bases dissociate completely in water, contributing all their hydroxide ions to the solution.

• The pOH of a strong base solution is determined by the concentration of hydroxide ions, [OH–], produced by the base.

• The contribution of water's autoionization to [OH–] is negligible in the presence of a strong base. ( but not negligible in a weak base)

• The metal cation from the base does not affect the acid–base properties of the solution.

The relationship between pH and pOH is given by:

$pH + pOH = 14$

Example: Determining [H+] and [OH–] in a Strong Base Solution

• Given: 0.025 mol/L Ba(OH)2(aq)

• Dissociation: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$

• Each mole of Ba(OH)2 yields 2 moles of OH–:

• $[OH^-] = 2 \times 0.025 = 0.050$ mol/L

• Use the ion-product constant for water, $K_w$:

$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{0.050} = 2.0 \times 10^{-13} \text{ mol/L}$

• Conclusion: [H+] = $2.0 \times 10^{-13}$ mol/L, [OH–] = 0.050 mol/L

Example: Calculating pH from Mass of Strong Base

• Given: 2.6 g NaOH dissolved in 500.0 mL water

• Molar mass NaOH = 40.00 g/mol

• Moles NaOH: $n = \frac{2.6}{40.00} = 0.065$ mol

• Concentration: $c = \frac{0.065}{0.500} = 0.13$ mol/L

• Since NaOH dissociates completely: [OH–] = 0.13 mol/L

• Calculate [H+]: $[H^+] = \frac{1.0 \times 10^{-14}}{0.13} = 7.7 \times 10^{-14}$ mol/L

• Calculate pH: $pH = -\log(7.7 \times 10^{-14}) = 13.11$

Practice Problems

• Calculate [H+] and [OH–] for 0.00100 mol/L KOH(aq): [OH–] = 0.00100 mol/L; [H+] = $1.00 \times 10^{-11}$ mol/L

• Calculate pH for 0.042 mol Sr(OH)2 in 2.00 L: pH = 3.92

Calculations Involving Solutions of Weak Bases

For weak bases, the base ionization constant (Kb) is used in calculations, analogous to the use of Ka for weak acids. If Kb is not given, it can be calculated from the K_a of the conjugate acid using:

$K_a K_b = K_w$

Example: Calculating Kb and pH for a Weak Base (Ammonia)

• Given: [NH3(aq)] = 15.0 mol/L

• Conjugate acid: NH4+(aq), $K_a = 5.8 \times 10^{-10}$

• Calculate $K_b$:

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{5.8 \times 10^{-10}} = 1.724 \times 10^{-5}$

• Set up the equilibrium (ICE) table for NH3 ionization:

	NH3(aq)	OH–(aq)	NH4+(aq)
Initial (I)	15.0	0	0
Change (C)	–x	+x	+x
Equilibrium (E)	15.0 – x	x	x

• Assume $15.0 - x \approx 15.0$ (since $\frac{[NH_3]_{initial}}{K_b} \gg 100$)

• Substitute into the equilibrium expression:

$K_b = \frac{x^2}{15.0}$

$x^2 = K_b \times 15.0 = 1.724 \times 10^{-5} \times 15.0 = 2.6 \times 10^{-4}$

$x = \sqrt{2.6 \times 10^{-4}} = 1.612 \times 10^{-2}$

• [OH–] = $1.612 \times 10^{-2}$ mol/L

• pOH = $-\log(1.612 \times 10^{-2}) = 1.79$

• pH = $14 - 1.79 = 12.21$

Statement: The pH of the ammonia solution is 12.21.

Practice Problems for Weak Bases

• Determine Kb for ethanoate ion, C2H3O2–(aq): $5.6 \times 10^{-10}$

• Determine Kb for borate ion, H2BO3–(aq): $1.7 \times 10^{-5}$

• Calculate pH for 0.20 mol/L base, $K_b = 3.82 \times 10^{-10}$: pH = 8.94

• Determine pH for 4.5 mol/L hydrazine, N2H4(aq): pH = 11.44

Summary

• Calculations for weak base solutions are analogous to those for weak acids.

• Ka values are often available in reference tables; Kb can be calculated using $K_a K_b = K_w$.

Additional info:

• For all calculations, ensure units are consistent (e.g., convert mL to L for concentration).

• When using the ICE table, check if the simplifying assumption ($C_{initial} \gg K_b$) is valid by the "hundred rule" ($\frac{C_{initial}}{K_b} > 100$).

• Always check the percent ionization (5% rule) to validate the assumption that $C_{initial} - x \approx C_{initial}$.

• My Note: Weak acids form an equilibrium, vs. strong acids that go to completion