Answer
The most acidic solution is (a) 0.10 HCl, since it has the lowest pH.
Work Step by Step
(a)
1. $HCl$ is a strong acid, therefore:
$[H_3O^+] = [HCl]_{initial} = 0.10 \space M$
2. $pH = -log([H_3O^+]) = -log(0.10) = 1.00$
(b) We do not really have to calculate this pH to know that a 0.10 M HF solution is less acidic than a 0.10 M HCl one, because HF is a weak acid and HCl is a strong acid.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\
Initial& 0.10 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.10 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
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 $ 0.10 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.10 }$$
$x = 8.2 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 8.2 \times 10^{-3} }{ 0.10 } \times 100\% = 8.2 \%$$
The percent is greater than 5%; therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$
$$K_a [ HF ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HF ] = 0$$
$$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.10 )} }{2 (1)}$$
$$x_1 = 7.9 \times 10^{-3} $$
$$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.10 )} }{2 (1)}$$
$$x_2 = -8.6 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 7.9 \times 10^{-3} $$
6. $$[H_3O^+] = x = 7.9 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 7.9 \times 10^{-3} ) = 2.10 $$
(c)
3. Assuming $ 0.20 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.20 }$$
$x = 0.012 $
4. Test if the assumption was correct:
$$\frac{ 0.012 }{ 0.20 } \times 100\% = 6.0 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HF ]_{initial} - x}$$
$$K_a [ HF ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HF ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 6.8 \times 10^{-4} + \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} ) ( 0.20 )} }{2 (1)}$$
$$x_1 = 0.011 $$
$$x_2 = \frac{- 6.8 \times 10^{-4} - \sqrt{( 6.8 \times 10^{-4} )^2 - 4 (1) (- 6.8 \times 10^{-4} )( 0.20 )} }{2 (1)}$$
$$x_2 = -0.012 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 0.011 $$
6. $$[H_3O^+] = x = 0.011 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.011 ) = 1.96 $$
