Answer
The percent dissociation is equal to 1.5%
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCOOH ]& [ HCOO^- ]& [ H_3O^+ ]\\
Initial& 0.75 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.75 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HCOO^- ][ H_3O^+ ]}{[ HCOOH ]}$$
$$K_a = \frac{(x)(x)}{[ HCOOH ]_{initial} - x}$$
3. Assuming $ 0.75 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCOOH ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCOOH ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.75 }$$
$x = 0.011\underline 6 $
4. Test if the assumption was correct:
$$\frac{ 0.011\underline 6 }{ 0.75 } \times 100\% = 1.5 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.011 $
