## Chemistry: Molecular Approach (4th Edition)

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\ Initial& 0.0075 & 0 & 0.0075 \\ Change& -x& +x& +x\\ Equilibrium& 0.0075 -x& 0 +x& 0.0075 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$ $$K_a = \frac{(x)( 0.0075 + x)}{[ HS{O_4}^- ]_{initial} - x}$$ 3. Solve for x: $$K_a = \frac{( 0.0075 + x)(x)}{[ HS{O_4}^- ]_{initial} - x}$$ $$K_a [ HS{O_4}^- ] - K_a x = 0.0075 x + x^2$$ $$x^2 + (K_a + 0.0075 ) x - K_a [ HS{O_4}^- ] = 0$$ $$x_1 = \frac{- ( 1.2 \times 10^{-2} + 0.0075 )+ \sqrt{( 1.2 \times 10^{-2} + 0.0075 )^2 - 4 (1) (- 1.2 \times 10^{-2} ) ( 0.0075 )} }{2 (1)}$$ $$x_1 = 3.854 \times 10^{-3}$$ $$x_2 = \frac{- ( 1.2 \times 10^{-2} + 0.0075 )- \sqrt{( 1.2 \times 10^{-2} + 0.0075 )^2 - 4 (1) (- 1.2 \times 10^{-2} ) ( 0.0075 )} }{2 (1)}$$ $$x_2 = -0.023$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = -0.023$$ 6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 0.0075 + 3.854 \times 10^{-3} = 0.011354$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.011354 ) = 1.945$$