Answer to Question #220371 in General Chemistry for gyuioh

Question

Calculate the pH of a 0.59M solution of sodium acetate (NaCH3COO)

Answer 1

Sodium acetate is a salt of a strong base and weak acid.

K_a of NaCH₃COO $= 1.08 \times 10^{-5}$

$K_a \times K_b = 10^{-14} \\ K_b = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

$K_b = 5.56 \times 10^{-10}= \frac{[CH_3COOH][NaOH]}{[CH_3COONa]} \\ = \frac{x \times x }{0.59-x} = \frac{x^2}{0.59-x}$

Since Kb is small, x <<<0.59

So, (0.59-x)≈0.59

$5.59 \times 10^{-10}= \frac{x^2}{0.59} \\ x = \sqrt{0.59 \times 5.56 \times 10^{-10}} = [OH^-] \\ [OH^-]=1.82 \times 10^{-5 } \\ pOH = -log[OH^-] = -log(1.82 \times 10^{-5})=4.74 \\ pH = 14-pOH \\ = 14-4.74 \\ = 9.26$

Answer: 9.26