## Chemistry: The Molecular Science (5th Edition)

$pH = 4.99$
1. Find the concentration after the dilution: $C_1 * V_1 = C_2 * V_2$ $0.1*1= C_2 *10$ $0.1 = C_2 *10$ $C_2 =0.01$ 2. Drawing the ICE table we get these concentrations at the equilibrium: $C_3H_5O_2H(aq) + H_2O(l) \lt -- \gt C_3H_5O{_2}^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[C_3H_5O_2H] = 0.2 M - x$ $[C_3H_5O{_2}^-] = 0.3M + x$ $[H_3O^+] = 0 + x$ 3. Calculate 'x' using the $K_a$ expression. $1.4\times 10^{- 5} = \frac{[C_3H_5O{_2}^-][H_3O^+]}{[C_3H_5O_2H]}$ $1.4\times 10^{- 5} = \frac{( 0.3 + x )* x}{ 0.2 - x}$ Considering 'x' has a very small value. $1.4\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$ $1.4\times 10^{- 5} = 1.5x$ $\frac{ 1.4\times 10^{- 5}}{ 1.5} = x$ $x = 9.333\times 10^{- 6}$ Percent dissociation: $\frac{ 9.333\times 10^{- 6}}{ 0.2} \times 100\% = 4.667\times 10^{- 3}\%$ x = $[H_3O^+]$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 9.333 \times 10^{- 6})$ $pH = 5.03$ 5. Since we are adding a strong acid, this reaction will occur: $C_3H_5O{_2}^-(aq) + H_3O^+(aq) -- \gt C_3H_5O_2H(aq) + H_2O(l)$ And these are the concentrations after this reaction: Remember: Reactants at equilibrium = Initial Concentration - y And Products = Initial Concentration + y Since $HCl$ is a strong acid, y = $[HCl] = 0.01M$ $[C_3H_5O_2H] = 0.2 M + 0.01 = 0.21M$ $[C_3H_5O{_2}^-] = 0.3M - 0.01 = 0.28M$ 6. Now, calculate the hydronium ion concentration after the addition of the $HCl$: $[H_3O^+] = Ka * (\frac{[C_3H_5O_2H]}{[C_3H_5O{_2}^-]})$ $[H_3O^+] = 1.4 \times 10^{-5} * \frac{0.21}{0.28}$ $[H_3O^+] = 1.4 \times 10^{-5} * 0.7241$ $[H_3O^+] = 1.014 \times 10^{-5}$ 7. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.014 \times 10^{- 5})$ $pH = 4.99$