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

(a) $$Pressure = 316 \space Torr. \times \frac{1 \space atm}{760 \space Torr.} = 0.416 \space atm$$ $$Temperature/K = 25.0 + 273.15 = 298.0 \space K$$ 1. According to the Ideal Gas Law: $$\frac{n}{V} = molarity = \frac{P}{RT} = \frac{( 0.416 \space atm )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 298.0 \space K)}$$ $$C = 1.70 \times 10^{-2} \space M$$ 2. Find the $K_b$ $$K_b = 10^{-pKb} = 3.7 \times 10^{-4}$$ - Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ B^- ]& [ HB ]& [ OH^- ]\\ Initial& 1.70 \times 10^{-2} & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 1.70 \times 10^{-2} -x& 0 +x& 0 +x\\ \end{vmatrix}$$ - Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HB ][ OH^- ]}{[ B^- ]}$$ $$K_b = \frac{(x)(x)}{[ B^- ]_{initial} - x}$$ 3. Assuming $1.70 \times 10^{-2} \gt\gt x$: $$K_b = \frac{x^2}{[ B^- ]_{initial}}$$ $$x = \sqrt{K_b \times [ B^- ]_{initial}} = \sqrt{ 3.7 \times 10^{-4} \times 1.70 \times 10^{-2} }$$ $x = 2.5 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 2.5 \times 10^{-3} }{ 1.70 \times 10^{-2} } \times 100\% = 15.0 \%$$ 5. Return for the original expression and solve for x: $$K_b = \frac{x^2}{[ B^- ]_{initial} - x}$$ $$K_b [ B^- ] - K_b x = x^2$$ $$x^2 + K_b x - K_b [ B^- ] = 0$$ - Using bhaskara: $$x_1 = \frac{- 3.7 \times 10^{-4} + \sqrt{( 3.7 \times 10^{-4} )^2 - 4 (1) (- 3.7 \times 10^{-4} ) ( 1.70 \times 10^{-2} )} }{2 (1)}$$ $$x_1 = 2.3 \times 10^{-3}$$ $$x_2 = \frac{- 3.7 \times 10^{-4} - \sqrt{( 3.7 \times 10^{-4} )^2 - 4 (1) (- 3.7 \times 10^{-4} )( 1.70 \times 10^{-2} )} }{2 (1)}$$ $$x_2 = -2.7 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 2.3 \times 10^{-3}$$ 6. $[OH^-] = x = 2.3 \times 10^{-3}$ 7. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 2.3 \times 10^{-3} } = 4.3 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 4.3 \times 10^{-12} ) = 11.37$$ (b) Since NaOH is a strong base: $$[OH^-] = 2.3 \times 10^{-3} \space M = [NaOH]$$ $NaOH$ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol $$0.500 \space L \space solution \times \frac{2.3 \times 10^{-3} \space mol \space NaOH}{1 \space L \space solution} \times \frac{40.00 \space g}{1 \space mol} \times \frac{1000 \space mg}{1 \space g} = 46 \space mg \space NaOH$$