Question

Enter your answer in the provided box. Determine the pH of a 0.16 M solution of sodium acetate (CH3COONa) at 25°C. (K, of acetic acid = 1.8 x 10-5.) pH = 8.

Answer 1

Sodium acetate is the salt of weak acid CH3COOH and strong base NaOH.

CH3COONa)

It will dissociate as follows in water

CHЗСОON@ag) > CН3СООаg) + Natag

The sodium cation will not affect the pH of the solution as it is from a strong base.

The acetate ion however is the conjugate base of an weak acid acetic acid. Hence, it will react with water as follows

CHЗCOO + HО0) — СHЗСООH (a) + Он а (ag)

The equilibrium constant for the above reaction is called Kb of acetate ion and is given by

eq Кь [CH3COО-19

Note that the concentrations are their equilibrium values in the solution.

Now, Kb of acetate ion can also be calculated from the Ka of its conjugate acid (acetic acid).

Ka and Kb of an acid and conjugate base pair are related as

14 K, x K= 10-

Given that Ka of acetic acid =

-5 1.8 x 10

10-14 К, 10-14 5.56 х 10-1о Ка 1.8 x 10-5

Given that the initial concentration of sodium acetate = 0.16 M

Hence, we can create the following ICE table.

	CHЗСОО 3	CHЗСООН 3	$[OH^-]$
Initial, M	0.16	0	0
Change, M	-x	+x	+x
Equilibrium , M	0.16-x	x	x

Hence, we can write the Kb expression as

$K_b = \frac{[CH_3COOH]_{eq}[OH^-]_{eq}}{[CH_3COO^-]_{eq}} = 5.56 \times 10^{-10} \\ \\ \Rightarrow \frac{x \times x}{0.16-x} = 5.56 \times 10^{-10}\\ \Rightarrow x^2 = (0.16 - x ) \times 5.56 \times 10^{-10} \\ \Rightarrow x \approx 9.43 \times 10^{-6}$

Hence, the equilibrium OH- concentration of the solution is

он s ОН 1eg — х 3D 9.43 х 10-6 М

We know that the [OH-] and [H3O+] concentration are related as

[OHegH30eg = 10-14 10-14 [H3Ole 42 10-61.06 x 10 M

pH is defined as the negative logarithm of the H3O+ concentration, Hence, pH of the solution can be calculated as

pH logH30+]eg =- log (1.06 x 10)8.97

Hence, the pH of the solution is 8.97 approximately.