## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 78

#### Answer

a. pH = 3.00 and Percent Ionization = 1.0% b. pH = 2.02 and Percent Ionization = 9.5% c. pH = 1.21 and Percent Ionization = 62%

#### Work Step by Step

a. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$ $$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$ 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HA ]_{initial}}$$ $$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-5} \times 0.100 }$$ $x = 1.0 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 1.0 \times 10^{-3} }{ 0.100 } \times 100\% = 1.0 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 1.0 x 10^{-3}$ 6. $$[H^+] = x = 1.0 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 1.0 \times 10^{-3} ) = 3.00$$ ----- b. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HA ]_{initial}}$$ $$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-3} \times 0.100 }$$ $x = 0.010$ 4. Test if the assumption was correct: $$\frac{ 0.010 }{ 0.100 } \times 100\% = 10.0 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HA ]_{initial} - x}$$ $$K_a [ HA ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HA ] = 0$$ $$x_1 = \frac{- 1.0 \times 10^{-3} + \sqrt{( 1.0 \times 10^{-3} )^2 - 4 (1) (- 1.0 \times 10^{-3} ) ( 0.100 )} }{2 (1)}$$ $$x_1 = 9.5 \times 10^{-3}$$ $$x_2 = \frac{- 1.0 \times 10^{-3} - \sqrt{( 1.0 \times 10^{-3} )^2 - 4 (1) (- 1.0 \times 10^{-3} )( 0.100 )} }{2 (1)}$$ $$x_2 = -0.011$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 9.5 \times 10^{-3}$$ 6. $$[H^+] = x = 9.5 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-3} ) = 2.02$$ 8. Calculate the correct percent ionization: $$\frac{9.5 \times 10^{-3}}{0.100} \times 100\% = 9.5\%$$ ---------------- c. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HA ]_{initial}}$$ $$x = \sqrt{K_a \times [ HA ]_{initial}} = \sqrt{ 1.0 \times 10^{-1} \times 0.100 }$$ $x = 0.10$ 4. Test if the assumption was correct: $$\frac{ 0.10 }{ 0.100 } \times 100\% = 100.0 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HA ]_{initial} - x}$$ $$K_a [ HA ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HA ] = 0$$ $$x_1 = \frac{- 1.0 \times 10^{-1} + \sqrt{( 1.0 \times 10^{-1} )^2 - 4 (1) (- 1.0 \times 10^{-1} ) ( 0.100 )} }{2 (1)}$$ $$x_1 = 0.062$$ $$x_2 = \frac{- 1.0 \times 10^{-1} - \sqrt{( 1.0 \times 10^{-1} )^2 - 4 (1) (- 1.0 \times 10^{-1} )( 0.100 )} }{2 (1)}$$ $$x_2 = -0.16$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.062$$ 6. $$[H^+] = x = 0.062$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.062 ) = 1.21$$ 8. Find the correct percent ionization: $$\frac{0.062}{0.100} = 62\%$$

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