Question 16.15: Calculate the concentration of OH^- in a 0.15 M solution of ...
Calculate the concentration of OH^{-} in a 0.15 M solution of NH_{3}.
Analyze We are given the concentration of a weak base and asked to determine the concentration of OH^{-}.
Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids—that is, write the chemical equation and tabulate initial and equilibrium concentrations.
Solve
The ionization reaction and equilibrium expression are:
K_{b} =\frac{[NH^{+}_{4}][OH^{-}]}{[NH_{3}] }= 1.8 × 10^{-5}
Ignoring the concentration of H_{2}O because it is not involved in the equilibrium expression, we see that the equilibrium concentrations are:
Inserting these quantities into the equilibrium expression gives:
K_{b} =\frac{[NH^{+}_{4}][OH^{-}]}{[NH_{3}] }=\frac{(x)(x)}{0.15 – x}= 1.8 × 10^{-5}Because K_{b} is small, the amount of NH_{3} that reacts with water is much smaller than the NH_{3} concentration, and so we can neglect x relative to 0.15 M. Then we have:
\frac{x²}{0.15}= 1.8 × 10^{-5}
x² = (0.15)(1.8 × 10^{-5}) = 2.7 × 10^{-6}
x = [NH^{+}_{4}] = [OH^{-}] = \sqrt{2.7 × 10^{-6}} = 1.6 × 10^{-3} M
Check The value obtained for x is only about 1% of the NH_{3} concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified.
Comment You may be asked to find the pH of a solution of a weak base. Once you have found [OH^{-}], you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH_{3} contains [OH^{-}] = 1.6 × 10^{-3} M. Thus, pOH = -log(1.6 × 10^{-3}) = 2.80, and pH =14.00 – 2.80 = 11.20. The pH of the solution is above 7 because we are dealing with a solution of a base.