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?
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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