Question

Sodium acetate (NaOAc) is the sodium salt of the conjugate base of acetic acid (HOAc). Although NaOAc is a strong electrolyte, the acetate ion itself (OAc−) is a weak base. Assume pKa = 4.76 for HOAc.

1) Calculate the pH of a 50.0 mM HOAc solution. Using a 5% criterion, explicitly check any approximations you make to ensure they are valid.

2) Calculate the pH of a 50.0 mM NaOAc solution. Again, check your assumptions.

Answer 1

Given pKa of Acetic acid = 4.76

K_a of acetic acid = 10^-pka = 10^-4.76 = 1.74x10^-5

CH3COOH -----> CH3COO^- + H⁺

I 50.0 mM 0 0

C -x -x -x

E 50.0-x x x

According to 5% rule let us take 50x10^-3 - x = 50.x10^-3

x = 9.33x10^-4

(Let us test 5% rule

=1.866 < 5

So we can neglect x in denomnator)

Concentration of [H⁺] = x = 9.33x10^-4 M

pH = -log[H⁺] = 9.33x10^-4

pH =-log[9.33x10^-4 ]

pH =3.03

2) CH3COO^- + H2O -----> CH3COOH + OH^-

I 50x10^{-3 - 0 0}

^{C -x - +x +x}

E 50x10^-3-x - x x

Now we have to find Kb of acetic acid =

We know Ka*Kb = Kw= 10^-14

We have Ka = 1.74x10^-5

Kb = Kw/1.74x10^-5

K_b= 5.75x10^-10

Let us write equilibrium constant

Accrding to 5% rule neglecting denominator x

x = 5.36x10^-6

Concentration of OH^- = 5.36x10^-6

pOH = -log[OH^-]= 5.36x10^-6

pOH = -log[5.36x10^-6]

pOH = 5.27

pH = 14-pOH = 14-5.27 = 8.73

pH = 8.73