Answer
(a) pH = 4.74
(b) pOH = 12.56
Work Step by Step
(a)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCN ]& [ CN^- ]& [ H_3O^+ ]\\
Initial& 0.55 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.55 -x&x&x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CN^- ][ H_3O^+ ]}{[ HCN ]}$$
$$K_a = \frac{(x)(x)}{[ HCN ]_{initial} - x}$$
3. Assuming $ 0.55 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCN ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCN ]_{initial}} = \sqrt{ 6.2 \times 10^{-10} \times 0.55 }$$
$x = 1.8 \times 10^{-5} $
4. Test if the assumption was correct:
$$\frac{ 1.8 \times 10^{-5} }{ 0.55 } \times 100\% = 3.3 \times 10^{-3} \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 1.8 \times 10^{-5} $
6. $$[H_3O^+] = x = 1.8 \times 10^{-5} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 1.8 \times 10^{-5} ) = 4.74 $$
(b)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HIO_3 ]& [ I{O_3}^- ]& [ H_3O^+ ]\\
Initial& 0.044 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.044 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ I{O_3}^- ][ H_3O^+ ]}{[ HIO_3 ]}$$
$$K_a = \frac{(x)(x)}{[ HIO_3 ]_{initial} - x}$$
3. Assuming $ 0.044 \gt\gt x:$
$$K_a = \frac{x^2}{[ HIO_3 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HIO_3 ]_{initial}} = \sqrt{ 0.16 \times 0.044 }$$
$x = 0.084 $
4. Test if the assumption was correct:
$$\frac{ 0.084 }{ 0.044 } \times 100\% = 190.0 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HIO_3 ]_{initial} - x}$$
$$K_a [ HIO_3 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HIO_3 ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 0.16 + \sqrt{( 0.16 )^2 - 4 (1) (- 0.16 ) ( 0.044 )} }{2 (1)}$$
$$x_1 = 0.036 $$
$$x_2 = \frac{- 0.16 - \sqrt{( 0.16 )^2 - 4 (1) (- 0.16 )( 0.044 )} }{2 (1)}$$
$$x_2 = -0.20 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.036 $$
6. $$[H_3O^+] = x = 0.036 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.036 ) = 1.44 $$
$$pOH = 14.00 - 1.44 = 12.56$$
