## Chemistry: The Central Science (13th Edition)

$$K_a = 5.3\times 10^{- 5}$$
1. Calculate the $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.68}$ $[H_3O^+] = 2.1 \times 10^{-3}$ Therefore : $x = 2.1\times 10^{- 3}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: $C_6H_5CH_2COOH(aq) + H_2O(l) \lt -- \gt C_6H_5CH_2COOH^-(aq) + H_3O^+(aq)$ \begin{matrix} & [C_6H_5CH_2COOH] & [H_3O^+] & [C_6H_5CH_2COOH^-] \\ Initial & 0.085 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.085 -x & x & x \end{matrix} 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][C_6H_5CH_2COOH^-]}{ [C_6H_5CH_2COOH]}$ $Ka = \frac{x^2}{0.085 - x}$ $Ka = \frac{( 2.1\times 10^{- 3})^2}{ 0.085- 2.1\times 10^{- 3}}$ $Ka = 5.3\times 10^{- 5}$