## General Chemistry: Principles and Modern Applications (10th Edition)

0.500 M HF: $[H_3O^+] = 0.018 \space M$ $[HF] = 0.482 \space M$ 0.100 M HCl and 0.500 M HF: $[H_3O^+] = 0.103 \space M$ $[HF] = 0.497 \space M$
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. $$[H_3O^+] = x = 0.018$$ $[HF] = 0.500 - 0.018 = 0.482 \space M$ --------- 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: $$[H_3O^+] = x + [H_3O^+]_{initial} = 3.3 \times 10^{-3} + 0.100 = 0.103 \space M$$ $[HF] = 0.500 - 3.3 \times 10^{-3} = 0.497 \space M$