## Chemistry: Molecular Approach (4th Edition)

Hydroxide ion concentration $[OH^-] = 1.2 \times 10^{-2}$; pH = 12.08.
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ CH_3NH_2 ]& [ CH_3N{H_3}^{+} ]& [ OH^- ]\\ Initial& 0.33 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.33 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ CH_3N{H_3}^{+} ][ H^+ ]}{[ CH_3NH_2 ]}$$ $$K_b = \frac{(x)(x)}{[ CH_3NH_2 ]_{initial} - x}$$ 3. Assuming $0.33 \gt\gt x$: $$K_b = \frac{x^2}{[ CH_3NH_2 ]_{initial}}$$ $$x = \sqrt{K_b \times [ CH_3NH_2 ]_{initial}} = \sqrt{ 4.4 \times 10^{-4} \times 0.33 }$$ $x = 0.012$ 4. Test if the assumption was correct: $$\frac{ 0.012 }{ 0.33 } \times 100\% = 3.6 \%$$ 5. Thus, it is correct to say that $x = 0.012$ 6. $[OH^-] = x = 0.012 = 1.2 \times 10^{-2} \space M$ 7. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.012 } = 8.3 \times 10^{-13} \space M$$ $$pH = -log[H_3O^+] = -log( 8.3 \times 10^{-13} ) = 12.08$$