Answer
a. pH = 1.88, and the percent ionization is equal to 5.1%
b. pH = 2.10, and the percent ionization is equal to 7.9%
c. pH = 2.26, and the percent ionization is equal to 11%
Work Step by Step
a.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\
Initial& 0.250 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.250 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ F^- ][ H^+ ]}{[ HF ]}$$
$$K_a = \frac{(x)(x)}{[ HF ]_{initial} - x}$$
3. Assuming $ 0.250 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.250 }$$
$x = 0.013 $
4. Test if the assumption was correct:
$$\frac{ 0.013 }{ 0.250 } \times 100\% = 5.2 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$
$$K_a [ HF ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HF ] = 0$$
$$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.250 )} }{2 (1)}$$
$$x_1 = 0.012703 $$
$$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.250 )} }{2 (1)}$$
$$x_2 = -0.013 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.012703 \approx 0.013 (significant \space figures) $$
6. $$[H_3O^+] = x = 0.013$$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.013) = 1.88 $$
8. Calculate the correct percent ionization:
$$\frac{ 0.012703}{0.250} \times 100\% = 5.1\%$$
b.
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.100 }$$
$x = 8.2 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 8.2 \times 10^{-3} }{ 0.100 } \times 100\% = 8.2 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$
$$K_a [ HF ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HF ] = 0$$
$$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.100 )} }{2 (1)}$$
$$x_1 = 7.9 \times 10^{-3} $$
$$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.100 )} }{2 (1)}$$
$$x_2 = -8.6 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 7.9 \times 10^{-3} $$
6. $$[H^+] = x = 7.9 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 7.9 \times 10^{-3} ) = 2.10 $$
8. Calculate the correct percent ionization:
$$\frac{7.9 \times 10^{-3} }{0.100} \times 100\% = 7.9\%$$
c.
3. Assuming $ 0.050 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.050 }$$
$x = 5.8 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 5.8 \times 10^{-3} }{ 0.050 } \times 100\% = 12 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$
$$K_a [ HF ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HF ] = 0$$
$$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.050 )} }{2 (1)}$$
$$x_1 = 5.5 \times 10^{-3} $$
$$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.050 )} }{2 (1)}$$
$$x_2 = -6.2 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 5.5 \times 10^{-3} $$
6. $$[H^+] = x = 5.5 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 5.5 \times 10^{-3} ) = 2.26 $$
8. Calculate the correct percent ionization:
$$\frac{5.5 \times 10^{-3} }{0.050} \times 100\% =11\%$$
