## Chemistry (4th Edition)

$pH = 4.85$
$CH_3COONa$: $Na^+$ is not an electrolyte, but $CH_3COO^-$ acts as a base: - Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its $K_b$ by using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_b = 5.6\times 10^{- 10}$ 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3COO^-(aq) + H_2O(l) \lt -- \gt CH_3COOH(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3COOH$ is : 0 M, and $[OH^-]$ $\approx$ 0 M -$[OH^-] = [CH_3COOH] = 0 + x = x$ -$[CH_3COO^-] = [CH_3COO^-]_{initial} - x$ For approximation, we are going to consider $[CH_3COO^-]_{initial} = [CH_3COO^-]$ 2. Now, use the $K_b$ value and equation to find the 'x' value. $Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$ $Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.36}$ $Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.36}$ $2.0 \times 10^{- 10} = x^2$ $x = 1.4 \times 10^{- 5}$ Percent ionization: $\frac{ 1.4 \times 10^{- 5}}{ 0.36} \times 100\% = 0.0039\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [CH_3COOH] = x = 1.4 \times 10^{- 5}M$ $[CH_3COO^-] \approx 0.36M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.4 \times 10^{- 5})$ $pH = 4.85$