## Chemistry: Molecular Approach (4th Edition)

1. $Na^+$ does not ionize, so we can ignore it. $C_2H_3{O_2}^-$ is the conjugate base of a weak acid; thus, it is a weak base. 2. Calculate its $K_b$: $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$ 3. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ C_2H_3O{_2}^- ]& [ HC_2H_3O_2 ]& [ OH^- ]\\ Initial& 0.250 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.250 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 4. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HC_2H_3O_2 ][ H^+ ]}{[ C_2H_3O{_2}^- ]}$$ $$K_b = \frac{(x)(x)}{[ C_2H_3O{_2}^- ]_{initial} - x}$$ 5. Assuming $0.250 \gt\gt x$: $$K_b = \frac{x^2}{[ C_2H_3O{_2}^- ]_{initial}}$$ $$x = \sqrt{K_b \times [ C_2H_3O{_2}^- ]_{initial}} = \sqrt{ 5.6 \times 10^{-10} \times 0.250 }$$ $x = 1.183 \times 10^{-5}$ 6. Test if the assumption was correct: $$\frac{ 1.183 \times 10^{-5} }{ 0.250 } \times 100\% = 4.7 \times 10^{-3} \%$$ 7. Thus, it is correct to say that $x = 1.183 \times 10^{-5}$ 8. $[OH^-] = x = 1.183 \times 10^{-5}$ 9. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.183 \times 10^{-5} } = 8.453 \times 10^{-10} \space M$$ $$pH = -log[H_3O^+] = -log( 8.453 \times 10^{-10} ) = 9.07$$