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

1. Draw the ICE table for the equilibrium of only 0.100 M CH3COOH: $$\begin{vmatrix} Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& x& x\\ \end{vmatrix}$$ 2. Now, we have to find an initial concentration of hydronium ion, given by the HCl, that makes $x = [CH_3COO^-] = 1.0 \times 10^{-4}$. Let's call that "y". $$\begin{vmatrix} Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & y \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& x& y + x\\ \end{vmatrix}$$ $$K_a = 1.8 \times 10^{-5} = \frac{x(y+x)}{0.100 - x}$$ We know that x = $1.0 \times 10^{-4}$ $$1.8 \times 10^{-5} = \frac{(1.0 \times 10^{-4})(y+1.0 \times 10^{-4})}{0.100 - 1.0 \times 10^{-4}} = \frac{(1.0 \times 10^{-4})(y + 1.0 \times 10^{-4})}{0.100}$$ 3. Solve for y: y = 0.0179 M Therefore, we need to start with $0.0179 M$ of $H_3O^+$ in order to get this concentration at equilibrium, which means 0.0179 M of HCl. $$1.00 \space L \times \frac{0.0179 \space mol \space HCl}{1 \space L} = 0.0179 \space mol \space HCl$$ $$0.0179 \space mol \space HCl \times \frac{1 \space L}{12 \space mol \space HCl} = 1.49 \times 10^{-3} \space L = 1.49 mL \approx 1.5 mL$$ $$1.5 \space mL \times \frac{1 \space drop}{0.050 \space mL} = 30 \space drops$$