## Chemistry: Molecular Approach (4th Edition)

$[H_3O^+] = 2.5 \times 10^{-3} \space M$ $pH = 2.59$
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_7H_5O_2 ]& [ C_7H_5O{_2}^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_7H_5O{_2}^- ][ H_3O^+ ]}{[ HC_7H_5O_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HC_7H_5O_2 ]_{initial} - x}$$ 3. Assuming $0.100 \gt\gt x$: $$K_a = \frac{x^2}{[ HC_7H_5O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_7H_5O_2 ]_{initial}} = \sqrt{ 6.5 \times 10^{-5} \times 0.100 }$$ x = $2.5 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 2.5 \times 10^{-3} }{ 0.100 } \times 100\% = 2.5 \% \lt 5\%$$ 5. Thus, it is correct to say that $x = 2.5 \times 10^{-3}$ 6. $$[H_3O^+] = x = 2.5 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( \sqrt{ 6.5 \times 10^{-5} \times 0.100 } ) = 2.59$$