Question

# Sodium acetate (NaOAc) is the sodium salt of the conjugate base of acetic acid (HOAc). Although...

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.

Given pKa of Acetic acid = 4.76

Ka 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

Kb= 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

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