Answer
The pH of this solution is equal to 10.47
Work Step by Step
- Calculate the molar mass:
$ C_9H_{13}N $ : ( 1.008 $\times$ 13 )+ ( 12.01 $\times$ 9 )+ ( 14.01 $\times$ 1 )= 135.20 g/mol
- Use the information as conversion factors to find the molarity of this solution:
$$\frac{ 225 mg \space C_9H_{13}N }{1 \space L \space solution} \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mol \space C_9H_{13}N }{ 135.20 \space g \space C_9H_{13}N } = 1.66 \times 10^{-3} \space M$$
- Calculate the $K_b$
$$K_b = 10^{-pK_b} = 6.31 \times 10^{-5}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_9H_{13}N ]& [ C_9H_{14}N^{mais} ]& [ OH^- ]\\
Initial& 1.66 \times 10^{-3} & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.66 \times 10^{-3} -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_9H_{14}N^{mais} ][ H^+ ]}{[ C_9H_{13}N ]}$$
$$K_b = \frac{(x)(x)}{[ C_9H_{13}N ]_{initial} - x}$$
3. Assuming $ 1.66 x 10^{-3} \gt\gt x$:
$$K_b = \frac{x^2}{[ C_9H_{13}N ]_{initial}}$$
$$x = \sqrt{K_b \times [ C_9H_{13}N ]_{initial}} = \sqrt{ 6.31 \times 10^{-5} \times 1.66 \times 10^{-3} }$$
$x = 3.24 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 3.24 \times 10^{-4} }{ 1.66 \times 10^{-3} } \times 100\% = 19.5 \%$$
5. Return for the original expression and solve for x:
$$K_b = \frac{x^2}{[ C_9H_{13}N ]_{initial} - x}$$
$$K_b [ C_9H_{13}N ] - K_b x = x^2$$
$$x^2 + K_b x - K_b [ C_9H_{13}N ] = 0$$
$$x_1 = \frac{- 6.31 \times 10^{-5} + \sqrt{( 6.31 \times 10^{-5} )^2 - 4 (1) (- 6.31 \times 10^{-5} ) ( 1.66 \times 10^{-3} )} }{2 (1)}$$
$$x_1 = 2.94 \times 10^{-4} $$
$$x_2 = \frac{- 6.31 \times 10^{-5} - \sqrt{( 6.31 \times 10^{-5} )^2 - 4 (1) (- 6.31 \times 10^{-5} )( 1.66 \times 10^{-3} )} }{2 (1)}$$
$$x_2 = -3.57 \times 10^{-4} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 2.94 \times 10^{-4} $$
6. $[OH^-] = x = 2.94 \times 10^{-4} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 2.94 \times 10^{-4} } = 3.4 \times 10^{-11} \space M$$
$$pH = -log[H_3O^+] = -log( 3.4 \times 10^{-11} ) = 10.47 $$
