## Chemistry: Molecular Approach (4th Edition)

(a) Since HCl is a strong acid: $[H_3O^+] = [HCl]_{initial}$ $$pH = -log(1.0) = 0.00$$ (b) 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\ Initial& 2.0 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 2.0 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 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 $2.0 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 2.0 }$$ $x = 0.037$ 4. Test if the assumption was correct: $$\frac{ 0.037 }{ 2.0 } \times 100\% = 1.9 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 0.037$ 6. $$[H_3O^+] = x = 0.037$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.037 ) = 1.43$$ (c) As we saw in example 16.10, the $K_a$ for HClO is negligible compared to the same for HF. Thus, we can neglect the contribution of HClO to the pH: 3. Assuming $1.0 \gt\gt x:$ $$K_a = \frac{x^2}{[ HF ]_{initial}}$$ $$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 1.0 }$$ $x = 0.026$ 4. Test if the assumption was correct: $$\frac{ 0.026 }{ 1.0 } \times 100\% = 2.6 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 0.026$ 6. $$[H_3O^+] = x = 0.026$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.026 ) = 1.59$$