Exercises
Exercise 1. Calculate Kb for NH3 if Ka for NH4+ is equal to 5.6 x 10-10.
Ka x Kb = Kw
\(\displaystyle K_b\;=\;\frac{1.0\times 10^{-14}}{5.6\times 10^{-10}}\;=\;\mathbf{1.8\times 10^{-5}}\)
Exercise 2. Write a balanced net ionic equation and the equilibrium constant expression for the weak base ethlyamine, C2H5NH2, in water. Ka (C2H5NH3+) = 8.6 x 10-4.
C2H5NH2 (aq) + H2O (l) ⇄ C2H5NH3+ (aq) + OH– (aq)
\(\displaystyle K_b\;=\;\frac{1.0\times 10^{-14}}{8.6\times 10^{-4}}\;=\;1.2\times 10^{-11}\)
\(\displaystyle K_b\;=\;\frac{[C_2H_5NH_3^+][OH^-]}{C_2H_5NH_2}\;=\;1.2\times 10^{-11}\)
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Exercise 3. Pyridine, C5H5N, is a base with Kb = 2.0 x 10-9. What is the pH of a 0.75 M pyridine solution?
C5H5N (aq) + H2O (l) ⇄ C5H5NH+ (aq) + OH- (aq)
\(\displaystyle K_b\;=\;\frac{[C_5H_5NH^+][OH^-]}{C_5H_5N}\;=\;2.0\times 10^{-9}\)
Set up an ICE table.
Assume x is negligible using the 5% rule. 0.75 M – x ≅ 0.75 M
\(\displaystyle \frac{x^2}{0.75}\;=\;2.0\times 10^{-9}\)
\(\displaystyle x\;=\;\sqrt{0.75\times\;(2.0\times 10^{-9})}\;=\;3.9\times 10^{-5}\;M\)
Check assumption using the 5% rule.
\(\displaystyle \frac{3.9\times 10^{-5}\;M}{0.75\;M}\times\;100\;=\;0.005%\)
0.005% < 5%, our assumption is valid.
x = [OH–] = 3.9 x 10-5 M
pH = 14.00 – (-log(3.9 x 10-5)) = 9.59
Exercise 4. Ammonia, NH3 has a Kb = 1.8 x 10-5. What is the pH of an aqueous solution that is 0.35 M NH3?
NH3 (aq) + H2O (l) ⇄ NH4+ (aq) + OH– (aq)
\(\displaystyle K_b\;=\;\frac{[NH_4^+][OH^-]}{NH_3}\;=\;1.8\times 10^{-5}\)
Set up an ICE table.
Assume x is negligible using the 5% rule. 0.35 M – x ≅ 0.35 M
\(\displaystyle \frac{x^2}{0.35}\;=\;1.8\times 10^{-5}\)
\(\displaystyle x\;=\;\sqrt{0.35\times\;(1.8\times 10^{-5})}\;=\;2.5\times 10^{-3}\;M\)
Check assumption using the 5% rule.
\(\displaystyle \frac{2.5\times 10^{-3}\;M}{0.35\;M}\times\;100\;=\;0.71%\)
0.71% < 5%, our assumption is valid.
x = [OH–] = 2.5 x 10-3 M
pH = 14.00 – (-log(2.5 x 10-3)) = 11.40
Exercise 5. A 0.75 M aqueous solution of methylamine, CH3NH2, has a pH of 12.26. Calculate Kb and the equilibrium concentrations of [CH3NH2], [CH3NH3+], and [OH–].
CH3NH2 (aq) + H2O (l) ⇄ CH3NH3+ (aq) + OH– (aq)
Set up an ICE table.
We can determine x from the pH. First determine the pOH.
pOH = 14.00 – 12.26 = 1.74
[OH–] = 10-1.74 = 1.8 x 10-2 M
x = 1.8 x 10-2 M
The equilibrium concentrations are:
[CH3NH2] = 0.75 M – x = 0.75 M – 1.8 x 10-2 M = 0.73 M
[CH3NH3+] = [OH–] = x = 1.8 x 10-2 M
\(\displaystyle K_b\;=\;\frac{(1.8\times 10^{-2})^2}{0.73}\;=\;4.4\times 10^{-4}\)
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