## Chemistry: Molecular Approach (4th Edition)

$$[H_3O^+] = 0.013 \space M$$
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. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ F^- ][ H_3O^+ ]}{[ HF ]}$$ $$K_a = \frac{(x)(x)}{[ HF ]_{initial} - x}$$ 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: $$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$ $$K_a [ HF ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HF ] = 0$$ $$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.250 )} }{2 (1)}$$ $$x_1 = 0.013$$ $$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.250 )} }{2 (1)}$$ $$x_2 = -0.013$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.013$$ 6. $$[H_3O^+] = x = 0.013$$