Answer
$pH = 1.80 $
$[OH^-] = 6.2 \times 10^{-13} \space M$
$pOH = 12.20$
$[H_3O^+] = 0.016 $
$[H_2C_3H_2O_4] = 0.184 \space M$
$[HC_3H_2{O_4}^{-}] = 0.016 \space M$
$[C_3H_2{O_4}^{2-}] = 2.0 \times 10^{-6}$
Work Step by Step
- First dissociation:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2C_3H_2O_4 ]& [ HC_3H_2{O_4}^- ]& [ H_3O^+ ]\\
Initial& 0.200 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.200 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HC_3H_2{O_4}^- ][ H_3O^+ ]}{[ H_2C_3H_2O_4 ]}$$
$$K_a = \frac{(x)(x)}{[ H_2C_3H_2O_4 ]_{initial} - x}$$
3. Assuming $ 0.200 \gt\gt x:$
$$K_a = \frac{x^2}{[ H_2C_3H_2O_4 ]_{initial}}$$
$$x = \sqrt{K_a \times [ H_2C_3H_2O_4 ]_{initial}} = \sqrt{ 1.4 \times 10^{-3} \times 0.200 }$$
$x = 0.017 $
4. Test if the assumption was correct:
$$\frac{ 0.017 }{ 0.200 } \times 100\% = 8.5 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ H_2C_3H_2O_4 ]_{initial} - x}$$
$$K_a [ H_2C_3H_2O_4 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ H_2C_3H_2O_4 ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 1.4 \times 10^{-3} + \sqrt{( 1.4 \times 10^{-3} )^2 - 4 (1) (- 1.4 \times 10^{-3} ) ( 0.200 )} }{2 (1)}$$
$$x_1 = 0.016 $$
$$x_2 = \frac{- 1.4 \times 10^{-3} - \sqrt{( 1.4 \times 10^{-3} )^2 - 4 (1) (- 1.4 \times 10^{-3} )( 0.200 )} }{2 (1)}$$
$$x_2 = -0.017 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.016 $$
$[H_2C_3H_2O_4] = 0.200 - 0.016 = 0.184 \space M$
-------
Second dissociation:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_3H_2{O_4}^- ]& [ C_3H_2{O_4}^{2-} ]& [ H_3O^+ ]\\
Initial& 0.016 & 0 & 0.016 \\
Change& -x& +x& +x\\
Equilibrium& 0.016 -x& 0 +x& 0.016 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ C_3H_2{O_4}^{2-} ][ H_3O^+ ]}{[ HC_3H_2{O_4}^- ]}$$
$$K_a = \frac{(x)( 0.016 + x)}{[ HC_3H_2{O_4}^- ]_{initial} - x}$$
3. Assuming $0.016 \gt\gt x:$
$$K_a = \frac{(x)( 0.016 )}{[ HC_3H_2{O_4}^- ]_{initial}}$$
$$x = \frac{K_a \times [ HC_3H_2{O_4}^- ]_{initial}}{ 0.016 } = \frac{ 2.0 \times 10^{-6} \times 0.016 }{ 0.016 } $$
$x = 2.0 \times 10^{-6} $
4. Test if the assumption was correct:
$$\frac{ 2.0 \times 10^{-6} }{ 0.016 } \times 100\% = 0.012 \%$$
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: $$[H_3O^+] = x + [H_3O^+]_{initial} = 2.0 \times 10^{-6} + 0.016 = 0.016 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.016 ) = 1.80 $$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.016 } = 6.2 \times 10^{-13} \space M$$
$$pOH = 14.00 - 1.80 = 12.20$$
$[HC_3H_2{O_4}^{-}] = 0.016 - 2.0 \times 10^{-6} = 0.016 \space M$
$[C_3H_2{O_4}^{2-}] = x = 2.0 \times 10^{-6}$
