Chemistry: Molecular Approach (4th Edition)

- 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: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 2.94 \times 10^{-4} } = 3.4 \times 10^{-11} \space M$$ $$pH = -log[H_3O^+] = -log( 3.4 \times 10^{-11} ) = 10.47$$