Answer
$$pH = 11.28 $$
Work Step by Step
$ C_5H_{11}N $ : ( 1.008 $\times$ 11 )+ ( 12.01 $\times$ 5 )+ ( 14.01 $\times$ 1 )= 85.15 g/mol
- Calculate the amount of moles:
$$ 0.114 \space g \times \frac{1 \space mol}{ 85.15 \space g} = 1.34 \times 10^{-3} \space mol$$
- Calculate the molarity:
$$ \frac{ 1.34 \times 10^{-3} \space mol}{ 0.315 \space L} = 4.25 \times 10^{-3} \space M $$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_5H_{11}N ]& [ C_5H_{11}NH^{+} ]& [ OH^- ]\\
Initial& 4.25 \times 10^{-3} & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 4.25 \times 10^{-3} -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_5H_{11}NH^{+} ][ OH^- ]}{[ C_5H_{11}N ]}$$
$$K_b = \frac{(x)(x)}{[ C_5H_{11}N ]_{initial} - x}$$
3. Assuming $ 4.25 \times 10^{-3} \gt\gt x$:
$$K_b = \frac{x^2}{[ C_5H_{11}N ]_{initial}}$$
$$x = \sqrt{K_b \times [ C_5H_{11}N ]_{initial}} = \sqrt{ 1.6 \times 10^{-3} \times 4.25 \times 10^{-3} }$$
$x = 2.6 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 2.6 \times 10^{-3} }{ 4.25 \times 10^{-3} } \times 100\% = 61.0 \%$$
5. Return for the original expression and solve for x:
$$K_b = \frac{x^2}{[ C_5H_{11}N ]_{initial} - x}$$
$$K_b [ C_5H_{11}N ] - K_b x = x^2$$
$$x^2 + K_b x - K_b [ C_5H_{11}N ] = 0$$
$$x_1 = \frac{- 1.6 \times 10^{-3} + \sqrt{( 1.6 \times 10^{-3} )^2 - 4 (1) (- 1.6 \times 10^{-3} ) ( 4.25 \times 10^{-3} )} }{2 (1)}$$
$$x_1 = 1.9 \times 10^{-3} $$
$$x_2 = \frac{- 1.6 \times 10^{-3} - \sqrt{( 1.6 \times 10^{-3} )^2 - 4 (1) (- 1.6 \times 10^{-3} )( 4.25 \times 10^{-3} )} }{2 (1)}$$
$$x_2 = -3.5 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 1.9 \times 10^{-3} $$
6. $[OH^-] = x = 1.9 \times 10^{-3} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.9 \times 10^{-3} } = 5.3 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 5.3 \times 10^{-12} ) = 11.28 $$
