Acetate butler is often used as a buffer solution for protein studies at acidic
ID: 480740 • Letter: A
Question
Acetate butler is often used as a buffer solution for protein studies at acidic pH (from 3.6 to 5.6). The dissociation reaction is: CH_3COOH doubleheadarrow CH_3COO^- + H^+ Here CH_3COOH (abbreviated as HOAc) and the acetate anion CH_3COO^-(abbreviated as OAc^+) are the conjugate acid-base pair. Suppose you prepared an acetate buffer containing 0.1 M of HOAc and 0.1 M of OAc-. What are the [H^+] and pH of this solution? 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? Suppose you add 0.01 M HCl to a solution containing 0.18 M of HOAc and 0.02 M of OAc^-. What are the [H^+] and the pH before the HCl is added, and after the HCl is added? Suppose you add 0.01 M HCl to a solution containing 0.02 M of HOAc and 0.18 M of OAc^-. What are the [H^+] and the pH before the HCl is added, and after the HCl is added?Explanation / Answer
In the following parts, the calculations are done for 1L of the buffer solution.
(a) Using Henderson Hasselbach equation:
pH = pKa + log([cb]/[a])
Here [cb] is conc. of conjugate base and [a] is the conc. of acid.
Putting values, we get:
pH = pKa + log(0.1/0.1) = pKa
For acetic acid, pKa = 4.75
Thus, pH = 4.75 and [H+] = 10-pKa = 1.77*10-5 M
(b)
Moles of acid initially = 0.1
Moles of conjugate base initially = 0.1
Moles of HCl added = 0.01
Total moles of acid finally = 0.1 + 0.01 = 0.11
Total moles of conjugate base finally = 0.1-0.01 = 0.09 moles
Thus, using above equation we get:
pH = 4.75 + log(0.09/0.11) = 4.66
(c)
Before adding acid:
pH = 4.75 + log(0.02/0.18) = 3.79
After adding acid:
pH = 4.75 + log( (0.02-0.01)/(0.18+0.01) ) = 3.47
(d)
Before adding acid:
pH = 4.75 + log(0.18/0.02) = 5.7
After adding acid:
pH = 4.75 + log( (0.18-0.01)/(0.02+0.01) ) = 5.50