Answer
$$pH = 2.29 $$
Work Step by Step
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:
$$pH = -log[H_3O^+] = -log( 5.1 \times 10^{-3} ) = 2.29 $$
