Answer
0.500 M HF:
$[H_3O^+] = 0.018 \space M$
$[HF] = 0.482 \space M$
0.100 M HCl and 0.500 M HF:
$[H_3O^+] = 0.103 \space M$
$[HF] = 0.497 \space M$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\
Initial& 0.500 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.500 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
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 $ 0.500 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.6 \times 10^{-4} \times 0.500 }$$
$x = 0.018 $
4. Test if the assumption was correct:
$$\frac{ 0.018 }{ 0.500 } \times 100\% = 3.6 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.018 $
6. $$[H_3O^+] = x = 0.018 $$
$[HF] = 0.500 - 0.018 = 0.482 \space M$
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HCl is a strong acid, so a 0.100 M HCl solution produces $0.100 \space M \space H_3O^+$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\
Initial& 0.500 & 0 & 0.100 \\
Change& -x& +x& +x\\
Equilibrium& 0.500 -x& 0 +x& 0.100 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ F^- ][ H_3O^+ ]}{[ HF ]}$$
$$K_a = \frac{(x)( 0.100 + x)}{[ HF ]_{initial} - x}$$
3. Assuming $ 0.500 \space and \space 0.100 \gt\gt x:$
$$K_a = \frac{(x)( 0.100 )}{[ HF ]_{initial}}$$
$$x = \frac{K_a \times [ HF ]_{initial}}{ 0.100 } = \frac{ 6.6 \times 10^{-4} \times 0.500 }{ 0.100 } $$
$x = 3.3 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 3.3 \times 10^{-3} }{ 0.500 } \times 100\% = 0.66 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 3.3 \times 10^{-3} $
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 3.3 \times 10^{-3} + 0.100 = 0.103 \space M$$
$[HF] = 0.500 - 3.3 \times 10^{-3} = 0.497 \space M$
