Answer
a. 1.3%
b. 1.9%
c. 4.2%
d. 5.8%
Work Step by Step
a.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\
Initial& 1.00 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.00 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H^+ ]}{[ HCHO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$
3. Assuming $ 1.00 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 1.00 }$$
$x = 0.013 $
4. Test if the assumption was correct:
$$\frac{ 0.013 }{ 1.00 } \times 100\% = 1.3 \%$$
b.
3. Assuming $ 0.500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.500 }$$
$x = 9.5 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 9.5 \times 10^{-3} }{ 0.500 } \times 100\% = 1.9 \%$$
c.
3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.100 }$$
$x = 4.2 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 4.2 \times 10^{-3} }{ 0.100 } \times 100\% = 4.2 \%$$
d.
3. Assuming $ 0.0500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.0500 }$$
$x = 3.0 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 3.0 \times 10^{-3} }{ 0.0500 } \times 100\% = 6.0 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial} - x}$$
$$K_a [ HCHO_2 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HCHO_2 ] = 0$$
$$x_1 = \frac{- 1.8 \times 10^{-4} + \sqrt{( 1.8 \times 10^{-4} )^2 - 4 (1) (- 1.8 \times 10^{-4} ) ( 0.0500 )} }{2 (1)}$$
$$x_1 = 2.9 \times 10^{-3} $$
$$x_2 = \frac{- 1.8 \times 10^{-4} - \sqrt{( 1.8 \times 10^{-4} )^2 - 4 (1) (- 1.8 \times 10^{-4} )( 0.0500 )} }{2 (1)}$$
$$x_2 = -3.1 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 2.9 \times 10^{-3} $$
6. $$\frac{2.9 \times 10^{-3}}{0.0500} \times 100\% = 5.8\%$$
