Answer
a. pH =1.82, and we can make the assumption 'x is small'.
b. pH = 2.18, and we cannot make the assumption: 'x is small'.
b. pH = 2.72, and we cannot make the assumption: 'x is small'.
Work Step by Step
a.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HNO_2 ]& [ NO{_2}^- ]& [ H_3O^+ ]\\
Initial& 0.500 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.500 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ NO{_2}^- ][ H^+ ]}{[ HNO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HNO_2 ]_{initial} - x}$$
3. Assuming $ 0.500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.500 }$$
$x = 0.015 $
4. Test if the assumption was correct:
$$\frac{ 0.015 }{ 0.500 } \times 100\% = 3.0 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.015 $
6. $$[H^+] = x = 0.015 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.015 ) = 1.82 $$
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b. 3. Assuming $ 0.100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.100 }$$
$x = 0.0068 $
4. Test if the assumption was correct:
$$\frac{ 0.0068 }{ 0.100 } \times 100\% = 6.8 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$
$$K_a [ HNO_2 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HNO_2 ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.100 )} }{2 (1)}$$
$$x_1 = 0.0066 $$
$$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.100 )} }{2 (1)}$$
$$x_2 = -0.0069 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.0066$$
6. $$[H_3O^+] = x = 0.0066 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.0066 ) = 2.18 $$
------
c. 3. Assuming $ 0.0100 \gt\gt x:$
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.0100 }$$
$x = 0.0021 $
4. Test if the assumption was correct:
$$\frac{ 0.0021 }{ 0.0100 } \times 100\% = 21 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$
$$K_a [ HNO_2 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HNO_2 ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.0100 )} }{2 (1)}$$
$$x_1 = 0.0019 $$
$$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.0100 )} }{2 (1)}$$
$$x_2 = -0.0024 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.0019 $$
6. $$[H_3O^+] = x = 0.0019 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.0019 ) = 2.72 $$
