## Chemistry: Molecular Approach (4th Edition)

a. - HBr is a strong acid that will completely ionize, producing 0.115 M of $H_3O^+$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.125 & 0 & 0.115 \\ Change& -x& +x& +x\\ Equilibrium& 0.125 -x& 0 +x& 0.115 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$ $$K_a = \frac{(x)( 0.115 + x)}{[ HCHO_2 ]_{initial} - x}$$ 3. Assuming $0.125 \space and \space 0.115 \gt\gt x:$ $$K_a = \frac{(x)( 0.115 )}{[ HCHO_2 ]_{initial}}$$ $$x = \frac{K_a \times [ HCHO_2 ]_{initial}}{ 0.115 } = \frac{ 1.8 \times 10^{-4} \times 0.125 }{ 0.115 }$$ $x = 2.0 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 2.0 \times 10^{-4} }{ 0.125 } \times 100\% = 0.16 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 2.0 \times 10^{-4}$ 6. $$[H_3O^+] = x + [H_3O^+]_{initial} = 2.0 \times 10^{-4} + 0.115 = 0.115$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log(0.115) = 0.939$$ --------- b. - $HNO_3$ is a strong acid that will completely ionize, producing 0.085 M of $H_3O^+$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HNO_2 ]& [ N{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.150 & 0 & 0.085 \\ Change& -x& +x& +x\\ Equilibrium& 0.150 -x& 0 +x& 0.085 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ N{O_2}^- ][ H_3O^+ ]}{[ HNO_2 ]}$$ $$K_a = \frac{(x)( 0.085 + x)}{[ HNO_2 ]_{initial} - x}$$ 3. Assuming $0.150 \space and \space 0.085 \gt\gt x:$ $$K_a = \frac{(x)( 0.085 )}{[ HNO_2 ]_{initial}}$$ $$x = \frac{K_a \times [ HNO_2 ]_{initial}}{ 0.085 } = \frac{ 4.6 \times 10^{-4} \times 0.150 }{ 0.085 }$$ $x = 8.1 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 8.1 \times 10^{-4} }{ 0.150 } \times 100\% = 0.54 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 8.1 \times 10^{-4}$ 6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 8.1 \times 10^{-4} + 0.085 = 0.086$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.086 ) = 1.07$$ ----------- c. - Formic acid $(HCHO_2)$ is the stronger of these two; start by calculating the hydronium concentration of a pure formic acid with this molarity: 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.185 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.185 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$ 3. Assuming $0.185 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.185 }$$ $x = 5.8 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 5.8 \times 10^{-3} }{ 0.185 } \times 100\% = 3.1 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 5.8 x 10^{-3}$ 6. $$[H_3O^+] = x = 5.8 \times 10^{-3}$$ - Now, calculate the hydronium concentration of the mixture when we add the weaker acid: 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.225 & 0 & 5.8 x 10^{-3} \\ Change& -x& +x& +x\\ Equilibrium& 0.225 -x& 0 +x& 5.8 x 10^{-3} +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H_3O^+ ]}{[ HC_2H_3O_2 ]}$$ $$K_a = \frac{(x)( 5.8 \times 10^{-3} + x)}{[ HC_2H_3O_2 ]_{initial} - x}$$ 3. Assuming $0.225 \space and \space 5.8 x 10^{-3} \gt\gt x:$ $$K_a = \frac{(x)( 5.8 \times 10^{-3} )}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \frac{K_a \times [ HC_2H_3O_2 ]_{initial}}{ 5.8 \times 10^{-3} } = \frac{ 1.8 \times 10^{-5} \times 0.225 }{ 5.8 \times 10^{-3} }$$ $x = 7.0 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 7.0 \times 10^{-4} }{ 0.225 } \times 100\% = 0.31 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 7.0 \times 10^{-4}$ 6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 7.0 \times 10^{-4} + 5.8 \times 10^{-3} = 6.5 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 6.5 \times 10^{-3} ) = 2.19$$ --------- d. - Since the Ka for hydrocyanic acid is very small compared to the same for acetic acid, we can ignore the presence of hydrocyanic acid in the mixture: 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_2H_3O2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.050 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.050 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H_3O^+ ]}{[ HC_2H_3O2 ]}$$ $$K_a = \frac{(x)(x)}{[ HC_2H_3O2 ]_{initial} - x}$$ 3. Assuming $0.050 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.050 }$$ $x = 9.5 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 9.5 \times 10^{-4} }{ 0.050 } \times 100\% = 1.9 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 9.5 x 10^{-4}$ 6. $$[H_3O^+] = x = 9.5 \times 10^{-4}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-4} ) = 3.02$$