Chemistry: Molecular Approach (4th Edition)

1. Find the amount of moles in the 15.0 mL sample: $HC_2H_3O_2$ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 60.05g/mol $$15.0 \space mL \times \frac{1.05 \space g}{1 \space mL} \times \frac{1 \space mol}{60.05 \space g} = 0.262 \space mol$$ 2. Find the concentration of acetic acid in the resulting solution: $$\frac{0.262 \space mol}{1.50 \space L} =0.175 \space M$$ 3. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.175 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.175 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 4. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H^+ ]}{[ HC_2H_3O_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HC_2H_3O_2 ]_{initial} - x}$$ 5. Assuming $0.175 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.175 }$$ $x = 1.8 \times 10^{-3}$ 6. Test if the assumption was correct: $$\frac{ 1.8 \times 10^{-3} }{ 0.175 } \times 100\% = 1.0 \%$$ Percent ionization > 5%: Valid 7. The percent is less than 5%. Thus, it is correct to say that $x = 1.8 x 10^{-3}$ 8. $$[H_3O^+] = x = 1.8 \times 10^{-3}$$ 9. Calculate the pH: $$pH = -log[H_3O^+] = -log( \sqrt{ 1.8 \times 10^{-5} \times 0.175 } ) = 2.75$$