## Chemistry: Molecular Approach (4th Edition)

(a) 1. $HCl$ is a strong acid, therefore: $[H_3O^+] = [HCl]_{initial} = 0.10 \space M$ 2. $pH = -log([H_3O^+]) = -log(0.10) = 1.00$ (b) We do not really have to calculate this pH to know that a 0.10 M HF solution is less acidic than a 0.10 M HCl one, because HF is a weak acid and HCl is a strong acid. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\ Initial& 0.10 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.10 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ F^- ][ H_3O^+ ]}{[ HF ]}$$ $$K_a = \frac{(x)(x)}{[ HF ]_{initial} - x}$$ 3. Assuming $0.10 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.10 }$$ $x = 8.2 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 8.2 \times 10^{-3} }{ 0.10 } \times 100\% = 8.2 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.10 )} }{2 (1)}$$ $$x_1 = 7.9 \times 10^{-3}$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.10 )} }{2 (1)}$$ $$x_2 = -8.6 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 7.9 \times 10^{-3}$$ 6. $$[H_3O^+] = x = 7.9 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 7.9 \times 10^{-3} ) = 2.10$$ (c) 3. Assuming $0.20 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.20 }$$ $x = 0.012$ 4. Test if the assumption was correct: $$\frac{ 0.012 }{ 0.20 } \times 100\% = 6.0 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ - Using bhaskara: $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.20 )} }{2 (1)}$$ $$x_1 = 0.011$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.20 )} }{2 (1)}$$ $$x_2 = -0.012$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.011$$ 6. $$[H_3O^+] = x = 0.011$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.011 ) = 1.96$$