## General Chemistry: Principles and Modern Applications (10th Edition)

$[H_3O^+] = 0.026 \space M$ $[HS{O_4}^-] = 0.014 \space M$ $[S{O_4}^{2-}] = 6.0 \times 10^{-3} \space M$
1. Draw the ICE table for the first and second reactions: $$\begin{vmatrix} Compound& [ H_2SO_4 ]& [ HS{O_4}^- ]& [ H_2O ]\\ Initial& 0.020 & 0 & 0 \\ Change& -0.020& +0.020& +0.020\\ Equilibrium& 0& 0.020& 0.020 \\ \end{vmatrix}$$ Since $H_2SO_4$ is a strong acid, it will react completely, so x = 0.020 $$\begin{vmatrix} Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\ Initial& 0.020 & 0 & 0.020 \\ Change& -x& +x& +x\\ Equilibrium& 0.020 -x& 0 +x& 0.020 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$ $$K_a = \frac{(x)( 0.020 + x)}{0.020 - x}$$ 3. Assuming $0.020 \gt\gt x:$ $$K_a = \frac{(x)( 0.020 )}{0.020}$$ $$x = \frac{K_a \times [ HS{O_4}^- ]_{initial}}{ 0.020 } = \frac{ 1.1 \times 10^{-2} \times 0.020 }{ 0.020 }$$ $x = 0.011$ 4. Test if the assumption was correct: $$\frac{ 0.011 }{ 0.020 } \times 100\% = 55.0 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{( 0.020 + x)(x)}{[ HS{O_4}^- ]_{initial} - x}$$ $$K_a [ HS{O_4}^- ] - K_a x = 0.020 x + x^2$$ $$x^2 + (K_a + 0.020 ) x - K_a [ HS{O_4}^- ] = 0$$ $$x_1 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )+ \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$ $$x_1 = 6.0 \times 10^{-3}$$ $$x_2 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )- \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$ $$x_2 = -0.037$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 6.0 \times 10^{-3}$$ 6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 6.0 \times 10^{-3} + 0.020 = 0.026$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.026 ) = 1.59$$ $[H_3O^+] = 0.026 \space M$ $[HS{O_4}^-] = 0.020 - x = 0.014 \space M$ $[S{O_4}^{2-}] = x = 6.0 \times 10^{-3} \space M$