## Chemistry: Molecular Approach (4th Edition)

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\ Initial& 1.00 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 1.00 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H^+ ]}{[ HC_2H_3O_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HC_2H_3O_2 ]_{initial} - x}$$ a. 3. Assuming $1.00 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 1.00 }$$ $x = 4.2 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 4.2 \times 10^{-3} }{ 1.00 } \times 100\% = 0.42 \%$$ b. 3. Assuming $0.500 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.500 }$$ $x = 3.0 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 3.0 \times 10^{-3} }{ 0.500 } \times 100\% = 0.60 \%$$ c. 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.100 }$$ $x = 1.3 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 1.3 \times 10^{-3} }{ 0.100 } \times 100\% = 1.3 \%$$ d. 3. Assuming $0.0500 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.0500 }$$ $x = 9.5 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 9.5 \times 10^{-4} }{ 0.0500 } \times 100\% = 1.9 \%$$