## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 65

#### Answer

a. pH =1.82, and we can make the assumption 'x is small'. b. pH = 2.18, and we cannot make the assumption: 'x is small'. b. pH = 2.72, and we cannot make the assumption: 'x is small'.

#### Work Step by Step

a. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HNO_2 ]& [ NO{_2}^- ]& [ H_3O^+ ]\\ Initial& 0.500 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.500 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ NO{_2}^- ][ H^+ ]}{[ HNO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HNO_2 ]_{initial} - x}$$ 3. Assuming $0.500 \gt\gt x:$ $$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.500 }$$ $x = 0.015$ 4. Test if the assumption was correct: $$\frac{ 0.015 }{ 0.500 } \times 100\% = 3.0 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 0.015$ 6. $$[H^+] = x = 0.015$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.015 ) = 1.82$$ ----- b. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.100 }$$ $x = 0.0068$ 4. Test if the assumption was correct: $$\frac{ 0.0068 }{ 0.100 } \times 100\% = 6.8 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$ $$K_a [ HNO_2 ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HNO_2 ] = 0$$ - Using bhaskara: $$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.100 )} }{2 (1)}$$ $$x_1 = 0.0066$$ $$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.100 )} }{2 (1)}$$ $$x_2 = -0.0069$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.0066$$ 6. $$[H_3O^+] = x = 0.0066$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.0066 ) = 2.18$$ ------ c. 3. Assuming $0.0100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.0100 }$$ $x = 0.0021$ 4. Test if the assumption was correct: $$\frac{ 0.0021 }{ 0.0100 } \times 100\% = 21 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$ $$K_a [ HNO_2 ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HNO_2 ] = 0$$ - Using bhaskara: $$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.0100 )} }{2 (1)}$$ $$x_1 = 0.0019$$ $$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.0100 )} }{2 (1)}$$ $$x_2 = -0.0024$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.0019$$ 6. $$[H_3O^+] = x = 0.0019$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.0019 ) = 2.72$$

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