Answer
$ [ HOOCCH_2COO^- ] = 0.037 \space M$
$ [ ^-OOCCH_2COO^- ] = 2.0 \times 10^{-6} \space M$
$[H_3O^+] = 0.037 \space M $
Work Step by Step
- Find the results of the first ionization:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HOOCCH_2COOH ]& [ HOOCCH_2COO^- ]& [ H_3O^+ ]\\
Initial& 1.00 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.00 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HOOCCH_2COO^- ][ H_3O^+ ]}{[ HOOCCH_2COOH ]}$$
$$K_a = \frac{(x)(x)}{[ HOOCCH_2COOH ]_{initial} - x}$$
3. Assuming $ 1.00 \gt\gt x:$
$$K_a = \frac{x^2}{[ HOOCCH_2COOH ]_{initial}}$$
$$x = \sqrt{K_a \times [ HOOCCH_2COOH ]_{initial}} = \sqrt{ 1.4 \times 10^{-3} \times 1.00 }$$
$x = 0.037 $
4. Test if the assumption was correct:
$$\frac{ 0.037 }{ 1.00 } \times 100\% = 3.7 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.037 $
6. $$[H_3O^+] = x = 0.037 $$
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- Now, do the same for the second one.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HOOCCH_2COO^- ]& [ ^-OOCCH_2COO^- ]& [ H_3O^+ ]\\
Initial& 0.037 & 0 & 0.037 \\
Change& -x& +x& +x\\
Equilibrium& 0.037 -x& 0 +x& 0.037 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ ^-OOCCH_2COO^- ][ H_3O^+ ]}{[ HOOCCH_2COO^- ]}$$
$$K_a = \frac{(x)( 0.037 + x)}{[ HOOCCH_2COO^- ]_{initial} - x}$$
3. Assuming $ 0.037 \space and \space 0.037 \gt\gt x:$
$$K_a = \frac{(x)( 0.037 )}{[ HOOCCH_2COO^- ]_{initial}}$$
$$x = \frac{K_a \times [ HOOCCH_2COO^- ]_{initial}}{ 0.037 } = \frac{ 2.0 \times 10^{-6} \times 0.037 }{ 0.037 } $$
$x = 2.0 \times 10^{-6} $
4. Test if the assumption was correct:
$$\frac{ 2.0 \times 10^{-6} }{ 0.037 } \times 100\% = 5.4 \times 10^{-3} \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 2.0 \times 10^{-6} $
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 2.0 \times 10^{-6} + 0.037 = 0.037 $$
$ [ HOOCCH_2COO^- ] = 0.037 - x = 0.037 \space M$
$ [ ^-OOCCH_2COO^- ] = x = 2.0 \times 10^{-6}$
$[H_3O^+] = 0.037 + x = 0.037 $
