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

$pH = 5.03$
1. 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 $[C_3H_5O_2H] = 0.2 M - x$ $[C_3H_5O_2] = 0.3M + x$ $[H_3O^+] = 0 + x$ 2. 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. $1.4\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$ $1.4\times 10^{- 5} = 1.5x$ $\frac{ 1.4\times 10^{- 5}}{ 1.5} = x$ $x = 9.3 \times 10^{- 6}$ Percent dissociation: $\frac{ 9.3 \times 10^{- 6}}{ 0.2} \times 100\% = 4.7\times 10^{- 3}\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 9.3 \times 10^{- 6})$ $pH = 5.03$