Answer
a. pH = 3.00 and Percent Ionization = 1.0%
b. pH = 2.02 and Percent Ionization = 9.5%
c. pH = 1.21 and Percent Ionization = 62%
Work Step by Step
a.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\
Initial& 0.100 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$
$$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HA ]_{initial}}$$
$$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-5} \times 0.100 }$$
$x = 1.0 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 1.0 \times 10^{-3} }{ 0.100 } \times 100\% = 1.0 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 1.0 x 10^{-3} $
6. $$[H^+] = x = 1.0 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 1.0 \times 10^{-3} ) = 3.00 $$
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b.
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HA ]_{initial}}$$
$$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-3} \times 0.100 }$$
$x = 0.010 $
4. Test if the assumption was correct:
$$\frac{ 0.010 }{ 0.100 } \times 100\% = 10.0 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HA ]_{initial} - x}$$
$$K_a [ HA ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HA ] = 0$$
$$x_1 = \frac{- 1.0 \times 10^{-3} + \sqrt{( 1.0 \times 10^{-3} )^2 - 4 (1) (- 1.0 \times 10^{-3} ) ( 0.100 )} }{2 (1)}$$
$$x_1 = 9.5 \times 10^{-3} $$
$$x_2 = \frac{- 1.0 \times 10^{-3} - \sqrt{( 1.0 \times 10^{-3} )^2 - 4 (1) (- 1.0 \times 10^{-3} )( 0.100 )} }{2 (1)}$$
$$x_2 = -0.011 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 9.5 \times 10^{-3} $$
6. $$[H^+] = x = 9.5 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-3} ) = 2.02 $$
8. Calculate the correct percent ionization:
$$\frac{9.5 \times 10^{-3}}{0.100} \times 100\% = 9.5\%$$
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c.
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HA ]_{initial}}$$
$$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-1} \times 0.100 }$$
$x = 0.10 $
4. Test if the assumption was correct:
$$\frac{ 0.10 }{ 0.100 } \times 100\% = 100.0 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HA ]_{initial} - x}$$
$$K_a [ HA ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HA ] = 0$$
$$x_1 = \frac{- 1.0 \times 10^{-1} + \sqrt{( 1.0 \times 10^{-1} )^2 - 4 (1) (- 1.0 \times 10^{-1} ) ( 0.100 )} }{2 (1)}$$
$$x_1 = 0.062 $$
$$x_2 = \frac{- 1.0 \times 10^{-1} - \sqrt{( 1.0 \times 10^{-1} )^2 - 4 (1) (- 1.0 \times 10^{-1} )( 0.100 )} }{2 (1)}$$
$$x_2 = -0.16 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.062 $$
6. $$[H^+] = x = 0.062 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.062 ) = 1.21 $$
8. Find the correct percent ionization:
$$\frac{0.062}{0.100} = 62\%$$
