## General Chemistry: Principles and Modern Applications (10th Edition)

$$pH = 11.28$$
$C_5H_{11}N$ : ( 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: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.9 \times 10^{-3} } = 5.3 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 5.3 \times 10^{-12} ) = 11.28$$