Acetate buffer is often used as a buffer solution for protein studies at acidic

Question

Acetate buffer is often used as a buffer solution for protein studies at acidic pH (from 3.6 to 5.6). The dissociation reaction is: CH3COOH <---> CH3COO- + H+

Here CH3COOH (abbreviated as HOAc) and the acetate anion CH3COO- (abbreviated as OAc-) are the conjugate acid-base pair.

a.) Suppose you prepared an acetate buffer containing 0.1M of HOAc and 0.1M of OAc-. What are the [H+] and pH of this solution?

b.) Now you add 0.01M HCl to your acetate buffer. What are the [H+] and the pH after the HCl is added and the mixture reaches equilibrium?

c.) Suppose you add 0.01M HCl to a solution containing 0.18M of HOAc and 0.02M of OAc-. What are the [H+] and the pH before the HCl is added, and after the HCl is added?

d.) Suppose you add 0.01M HCl to a solution containing 0.02M of HOAc and 0.18M of OAc-. What are the [H+] and the pH before the HCl is added, and after the HCl is added?

Explanation / Answer

Acetate buffer is often used as a buffer solution for protein studies at acidic pH (from 3.6 to 5.6). The dissociation reaction is: CH3COOH <---> CH3COO- + H+

Here CH3COOH (abbreviated as HOAc) and the acetate anion CH3COO- (abbreviated as OAc-) are the conjugate acid-base pair.

We use Henderson equation to calculate pH

pH = pka + log ([Base]/ [acid])

pH = -log (1.75E-5) + log ( 0.1/0.1)

pH = 4.76

Solution :

Lets assume volume of buffer is 1.0 L ,

Mol calculate moles of HCl = 0.01 M * 1.0 L = 0.01 mol HCl

Mol OAc- = 0.1 M * 1.0 L = 0.1 mol

Mol HOAc = 0.1 M * 1.0 L = 0.1 mol

     OAc- + HCl   --> HOAc + Cl-

I   0.1         0.01           0.1          0

C -0.01      - 0.01      +0.01    +      0.01

E 0.09             0         0.11      

Now molarity of AcO- = 0.09/1.0 = 0.09 M

[HOAc]= 0.11/ 1.0 = 0.11 M

pH = 4.74 + log (0.09/0.11)

= 4.65

From pH we get H+

pH = -log (H+)

       [H+] = antilog ( -pH )

= antilog ( -4.65)

=2.22 E -5 M

So pH = 4.65 and [H+] = 2.22 E-5 M

Solution :

We Use Henderson equation to get pH before HCl is added

pH = 4.74 + log ( 0.02 / 0.18 )

= 4.04

[H+] = antilog ( -4.04) = 9.12 E-05

[H+]= 9.12011E-05

After addition of HCl

Mol HOAc = 0.18 M * 1.0 L = 0.18 mol

Mol AcO- = 0.020 M * 1.0 L = 0.020 mol

Now reaction of HCl with AcO-

Number of moles of HCl = 0.01 M* 1.0 L = 0.01 mol HCl

        OAc- + HCl -->HOAc + Cl-

I      0.020    0.01            0.18         0

C    -0.01     0.01            +0.01   

E    0.01        0                0.19

[HOAc] = 0.19 mol/ 1.0 L = 0.19 M

[AcO-] = 0.01/1.0 L = 0.01 M

pH = 4.74 + log (0.01/0.19)

=3.46

H+ = Antilog (-3.46) = 0.000346 M

Mole HCl = 0.01 M * 1.0 L = 0.01 mol HCl

pH = 4.74 +log ( 0.18/0.02)

= 5.42

[H+] = antilog ( -5.42) = 3.80189E-06

[H+]= 3.802E-06 M

After addition

Mol AcO-   = 0.18 M * 1.0 = 0.18mol , mol HOAc = 0.02 M *1.0 = 0.02

        OAc- + HCl ---> HOAc + Cl-

I        0.18    0.01            0.02         0

C    -0.01     0.01            +0.01   

E    0.17        0                0.03

[AcO-]= 0.17 mol/ 1.0 L = 0.17 M

[HOAc ] = 0.03 mol / 1.0 L = 0.03 M

pH = 4.74 + log ( 0.17 / 0.03)

=5.49

pH = 5.49

[H+]= antilog ( -5.49)

=3.23 E -6 M

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