Answer
$[H_3O^+] = 1.2 \times 10^{-4} \space M$
$[HCOO^-] =0.150 \space M$
Work Step by Step
$[HCOO^-] = [NaHCOO] = 0.150 M$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCOOH ]& [ HCOO^- ]& [ H_3O^+ ]\\
Initial& 0.100 & 0.150 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& 0.150 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HCOO^- ][ H_3O^+ ]}{[ HCOOH ]}$$
$$K_a = \frac{( 0.150 + x)(x)}{[ HCOOH ]_{initial} - x}$$
3. Assuming $ 0.100 \space and \space 0.150 \gt\gt x:$
$$K_a = \frac{( 0.150 )(x)}{[ HCOOH ]_{initial}}$$
$$x = \frac{K_a \times [ HCOOH ]_{initial}}{ 0.150 } = \frac{ 1.8 \times 10^{-4} \times 0.100 }{ 0.150 } $$
$x = 1.2 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 1.2 \times 10^{-4} }{ 0.100 } \times 100\% = 0.12 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 1.2 \times 10^{-4} $
6. $$[H_3O^+] = x = 1.2 \times 10^{-4} $$
$[HCOO^-] = 0.150 + 1.2 \times 10^{-4} = 0.150 \space M$
