Answer
The pH of this solution is equal to 1.945
Work Step by Step
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:
$$pH = -log[H_3O^+] = -log( 0.011354 ) = 1.945 $$
