## General Chemistry: Principles and Modern Applications (10th Edition)

$$pH = 2.29$$
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ CH_2FCOOH ]& [ CH_2FCOO^- ]& [ H_3O^+ ]\\ Initial& 0.015 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.015 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH_2FCOO^- ][ H_3O^+ ]}{[ CH_2FCOOH ]}$$ $$K_a = \frac{(x)(x)}{[ CH_2FCOOH ]_{initial} - x}$$ 3. Assuming $0.015 \gt\gt x:$ $$K_a = \frac{x^2}{[ CH_2FCOOH ]_{initial}}$$ $$x = \sqrt{K_a \times [ CH_2FCOOH ]_{initial}} = \sqrt{ 2.6 \times 10^{-3} \times 0.015 }$$ $x = 6.2 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 6.2 \times 10^{-3} }{ 0.015 } \times 100\% = 41.0 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ CH_2FCOOH ]_{initial} - x}$$ $$K_a [ CH_2FCOOH ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ CH_2FCOOH ] = 0$$ $$x_1 = \frac{- 2.6 \times 10^{-3} + \sqrt{( 2.6 \times 10^{-3} )^2 - 4 (1) (- 2.6 \times 10^{-3} ) ( 0.015 )} }{2 (1)}$$ $$x_1 = 5.1 \times 10^{-3}$$ $$x_2 = \frac{- 2.6 \times 10^{-3} - \sqrt{( 2.6 \times 10^{-3} )^2 - 4 (1) (- 2.6 \times 10^{-3} )( 0.015 )} }{2 (1)}$$ $$x_2 = -7.7 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 5.1 \times 10^{-3}$$ 6. $$[H_3O^+] = x = 5.1 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 5.1 \times 10^{-3} ) = 2.29$$