Answer
a. The percent ionization is equal to 0.42%
b. The percent ionization is equal to 0.60%
c. The percent ionization is equal to 1.3%
d. The percent ionization is equal to 1.9%
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\
Initial& 1.00 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.00 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H^+ ]}{[ HC_2H_3O_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_2H_3O_2 ]_{initial} - x}$$
a.
3. Assuming $ 1.00 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 1.00 }$$
$x = 4.2 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 4.2 \times 10^{-3} }{ 1.00 } \times 100\% = 0.42 \%$$
b.
3. Assuming $ 0.500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.500 }$$
$x = 3.0 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 3.0 \times 10^{-3} }{ 0.500 } \times 100\% = 0.60 \%$$
c.
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.100 }$$
$x = 1.3 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 1.3 \times 10^{-3} }{ 0.100 } \times 100\% = 1.3 \%$$
d.
3. Assuming $ 0.0500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.0500 }$$
$x = 9.5 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 9.5 \times 10^{-4} }{ 0.0500 } \times 100\% = 1.9 \%$$
