Answer
The pH of this solution is equal to 2.72
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HNO_2 ]& [ N{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.010 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.010 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ N{O_2}^- ][ H_3O^+ ]}{[ HNO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HNO_2 ]_{initial} - x}$$
3. Assuming $ 0.010 \gt\gt x:$
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.010 }$$
$x = 2.1 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 2.1 \times 10^{-3} }{ 0.010 } \times 100\% = 21.0 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$
$$K_a [ HNO_2 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HNO_2 ] = 0$$
$$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.010 )} }{2 (1)}$$
$$x_1 = 1.9 \times 10^{-3} $$
$$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.010 )} }{2 (1)}$$
$$x_2 = -2.4 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 1.9 \times 10^{-3} $$
6. $$[H_3O^+] = x = 1.9 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 1.9 \times 10^{-3} ) = 2.72 $$
