Answer
(a) 1.0 M HCl is the most acidic solution.
Work Step by Step
(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:
$$pH = -log[H_3O^+] = -log( 0.026 ) = 1.59 $$
