## Chemistry: Molecular Approach (4th Edition)

a. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\ Initial& 1.00 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 1.00 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H^+ ]}{[ HCHO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$ 3. Assuming $1.00 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 1.00 }$$ $x = 0.013$ 4. Test if the assumption was correct: $$\frac{ 0.013 }{ 1.00 } \times 100\% = 1.3 \%$$ b. 3. Assuming $0.500 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.500 }$$ $x = 9.5 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 9.5 \times 10^{-3} }{ 0.500 } \times 100\% = 1.9 \%$$ c. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.100 }$$ $x = 4.2 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 4.2 \times 10^{-3} }{ 0.100 } \times 100\% = 4.2 \%$$ d. 3. Assuming $0.0500 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.0500 }$$ $x = 3.0 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 3.0 \times 10^{-3} }{ 0.0500 } \times 100\% = 6.0 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial} - x}$$ $$K_a [ HCHO_2 ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HCHO_2 ] = 0$$ $$x_1 = \frac{- 1.8 \times 10^{-4} + \sqrt{( 1.8 \times 10^{-4} )^2 - 4 (1) (- 1.8 \times 10^{-4} ) ( 0.0500 )} }{2 (1)}$$ $$x_1 = 2.9 \times 10^{-3}$$ $$x_2 = \frac{- 1.8 \times 10^{-4} - \sqrt{( 1.8 \times 10^{-4} )^2 - 4 (1) (- 1.8 \times 10^{-4} )( 0.0500 )} }{2 (1)}$$ $$x_2 = -3.1 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 2.9 \times 10^{-3}$$ 6. $$\frac{2.9 \times 10^{-3}}{0.0500} \times 100\% = 5.8\%$$