Answer
(a) pH = 2.290
(b) pOH = 12.699
Work Step by Step
(a)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HY ]& [ Y^- ]& [ H_3O^+ ]\\
Initial& 0.175 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.175 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ Y^- ][ H_3O^+ ]}{[ HY ]}$$
$$K_a = \frac{(x)(x)}{[ HY ]_{initial} - x}$$
3. Assuming $ 0.175 \gt\gt x:$
$$K_a = \frac{x^2}{[ HY ]_{initial}}$$
$$x = \sqrt{K_a \times [ HY ]_{initial}} = \sqrt{ 1.50 \times 10^{-4} \times 0.175 }$$
$x = 5.123 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 5.12 \times 10^{-3} }{ 0.175 } \times 100\% = 2.93 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 5.123 \times 10^{-3} $
6. $$[H_3O^+] = x = 5.123 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 5.123 \times 10^{-3} ) = 2.290 $$
(b)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HX ]& [ X^- ]& [ H_3O^+ ]\\
Initial& 0.175 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.175 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ X^- ][ H_3O^+ ]}{[ HX ]}$$
$$K_a = \frac{(x)(x)}{[ HX ]_{initial} - x}$$
3. Assuming $ 0.175 \gt\gt x:$
$$K_a = \frac{x^2}{[ HX ]_{initial}}$$
$$x = \sqrt{K_a \times [ HX ]_{initial}} = \sqrt{ 2.00 \times 10^{-2} \times 0.175 }$$
$x = 0.0592 $
4. Test if the assumption was correct:
$$\frac{ 0.0592 }{ 0.175 } \times 100\% = 33.8 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HX ]_{initial} - x}$$
$$K_a [ HX ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HX ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 2.00 \times 10^{-2} + \sqrt{( 2.00 \times 10^{-2} )^2 - 4 (1) (- 2.00 \times 10^{-2} ) ( 0.175 )} }{2 (1)}$$
$$x_1 = 0.0500 $$
$$x_2 = \frac{- 2.00 \times 10^{-2} - \sqrt{( 2.00 \times 10^{-2} )^2 - 4 (1) (- 2.00 \times 10^{-2} )( 0.175 )} }{2 (1)}$$
$$x_2 = -0.0700 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.0500 $$
6. $$[H_3O^+] = x = 0.0500 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.0500 ) = 1.301 $$
$$pOH = 14.00 - 1.301 = 12.699$$
