What is the molarity of a sodium acetate solution with a pH of 9.54?

To find the molarity of the sodium acetate solution (NaCH₃COO) with a pH of 9.54, we start by calculating the concentration of hydroxide ions (^OH–) using the pH value.

The pOH can be calculated as follows:

pOH = 14 – pH = 14 – 9.54 = 4.46

Now, we can find the concentration of hydroxide ions:

[^OH–] = 10^-pOH = 10^-4.46 ≈ 3.47 x 10^-5 M

Next, we convert to pK_b to find the dissociation of base; since sodium acetate is the salt of acetic acid (a weak acid), we use the relationship between the acid and base dissociation constants:

pK_b = 14 – pK_a = 14 – (-log(1.8 x 10^-5)) = 14 – 4.74 = 9.26

Now we can calculate K_b using:

K_b = 10^-pK_b = 10^-9.26 ≈ 4.74 x 10^-10

The equilibrium expression for the dissociation of sodium acetate is:

NaCH₃COO <=> Na⁺ + CH₃COO^–

At this point, we can use the following relationship involving K_b, OH^–, and the base concentration (C):

K_b = [^OH–][CH₃COO^–]/[NaCH₃COO]

Since we assume the complete dissolution of sodium acetate, the concentration of acetate ions [CH₃COO^–] is approximately equal to the initial concentration C. Thus, we assume C = x:

K_b = (3.47 x 10^-5)² / (C – 3.47 x 10^-5)

Neglecting 3.47 x 10^-5 compared to C (as it is quite small), we rewrite the equation as follows:

4.74 x 10^-10 = (3.47 x 10^-5)² / C

Rearranging gives us:

C = (3.47 x 10^-5)² / (4.74 x 10^-10)

Calculating the right side:

C ≈ 2.55 x 10^-3 M

Therefore, the molarity of the sodium acetate solution is approximately 2.55 mM.