## Chemistry: Molecular Approach (4th Edition)

$[OH^-] = 1.6 \times 10^{-3} \space M$ pH = 11.21 pOH = 2.79
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ NH_3 ]& [ NH{_4}^{+} ]& [ OH^- ]\\ Initial& 0.15 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.15 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ NH{_4}^{+} ][ H^+ ]}{[ NH_3 ]}$$ $$K_b = \frac{(x)(x)}{[ NH_3 ]_{initial} - x}$$ 3. Assuming $0.15 \gt\gt x$: $$K_b = \frac{x^2}{[ NH_3 ]_{initial}}$$ $$x = \sqrt{K_b \times [ NH_3 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.15 }$$ $x = 1.6 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 1.6 \times 10^{-3} }{ 0.15 } \times 100\% = 1.1 \%$$ 5. Thus, it is correct to say that $x = 1.6 \times 10^{-3}$ 6. $[OH^-] = x = 1.6 \times 10^{-3}$ 7. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.6 \times 10^{-3} } = 6.2 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 6.2 \times 10^{-12} ) = 11.21$$ $$pOH = 14 - pH = 14 - 11.21 = 2.79$$