## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 77

#### Answer

a. pH = 1.88, and the percent ionization is equal to 5.1% b. pH = 2.10, and the percent ionization is equal to 7.9% c. pH = 2.26, and the percent ionization is equal to 11%

#### Work Step by Step

a. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\ Initial& 0.250 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.250 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ F^- ][ H^+ ]}{[ HF ]}$$ $$K_a = \frac{(x)(x)}{[ HF ]_{initial} - x}$$ 3. Assuming $0.250 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.250 }$$ $x = 0.013$ 4. Test if the assumption was correct: $$\frac{ 0.013 }{ 0.250 } \times 100\% = 5.2 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.250 )} }{2 (1)}$$ $$x_1 = 0.012703$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.250 )} }{2 (1)}$$ $$x_2 = -0.013$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.012703 \approx 0.013 (significant \space figures)$$ 6. $$[H_3O^+] = x = 0.013$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.013) = 1.88$$ 8. Calculate the correct percent ionization: $$\frac{ 0.012703}{0.250} \times 100\% = 5.1\%$$ b. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.100 }$$ $x = 8.2 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 8.2 \times 10^{-3} }{ 0.100 } \times 100\% = 8.2 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.100 )} }{2 (1)}$$ $$x_1 = 7.9 \times 10^{-3}$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.100 )} }{2 (1)}$$ $$x_2 = -8.6 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 7.9 \times 10^{-3}$$ 6. $$[H^+] = x = 7.9 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 7.9 \times 10^{-3} ) = 2.10$$ 8. Calculate the correct percent ionization: $$\frac{7.9 \times 10^{-3} }{0.100} \times 100\% = 7.9\%$$ c. 3. Assuming $0.050 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.050 }$$ $x = 5.8 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 5.8 \times 10^{-3} }{ 0.050 } \times 100\% = 12 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.050 )} }{2 (1)}$$ $$x_1 = 5.5 \times 10^{-3}$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.050 )} }{2 (1)}$$ $$x_2 = -6.2 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 5.5 \times 10^{-3}$$ 6. $$[H^+] = x = 5.5 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 5.5 \times 10^{-3} ) = 2.26$$ 8. Calculate the correct percent ionization: $$\frac{5.5 \times 10^{-3} }{0.050} \times 100\% =11\%$$

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