1) Acetate buffer is often used as a buffer solution for proteinstudies at acidi
ID: 677577 • Letter: 1
Question
1) Acetate buffer is often used as a buffer solution for proteinstudies at acidic pH (from 3.6 to 5.6).
The dissociation reaction is: CH3COOH CH3COO- +H+
Here CH3COOH (abbreviated as HOAc) and the acetate anionCH3COO-(abbreviated as OAc-) are
the conjugate acid-base pair.
(a) Suppose you prepared an acetate buffer containing 0.1 Mof HOAc and 0.1 M of OAc-.
What are the [H+] and pH of this solution?
(b) Now you add 0.01M HCl to your acetate buffer. What arethe [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 0.18 Mof HOAc and 0.02 M of
OAc-. What are the [H+] and the pH before the HCl is added, andafter the HCl is added?
(d) Suppose you add 0.01M HCl to a solution containing 0.02 Mof HOAc and 0.18 M of
OAc-. What are the [H+] and the pH before the HCl is added, andafter the HCl is added?
1) Acetate buffer is often used as a buffer solution for proteinstudies at acidic pH (from 3.6 to 5.6).
The dissociation reaction is: CH3COOH CH3COO- +H+
Here CH3COOH (abbreviated as HOAc) and the acetate anionCH3COO-(abbreviated as OAc-) are
the conjugate acid-base pair.
(a) Suppose you prepared an acetate buffer containing 0.1 Mof HOAc and 0.1 M of OAc-.
What are the [H+] and pH of this solution?
(b) Now you add 0.01M HCl to your acetate buffer. What arethe [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 0.18 Mof HOAc and 0.02 M of
OAc-. What are the [H+] and the pH before the HCl is added, andafter the HCl is added?
(d) Suppose you add 0.01M HCl to a solution containing 0.02 Mof HOAc and 0.18 M of
OAc-. What are the [H+] and the pH before the HCl is added, andafter the HCl is added?
Explanation / Answer
Formula : pH = pKa + log [A-]/[HA] a ) pKa for Acetate buffer = 4.7447 pH = 4.7447 + log ( 0.1 / 0.1 ) pH = 4.7447 [H+] = 1.8 * 10 ^ -5M b) Let the total volume of the buffer solution is 1L and tothat 10 mL of 0.01 M HCl is added . Then no.of mols of salt decreases and no.of mols of acidincreases when a Acid is added to the acidic buffer . No.of mols of HCl = 0.0001mol No.of mols of salt = 0.1 mol - 0.0001 mol = 0.0999 mol No.of mols of acid = 0.1 mol + 0.0001 mol = 0.1001 mol pH = 4.7447 + log ( 0.0999 / 0.1001 ) =4.7438 [H+]= 1.8 * 10 ^ -5 M c) pH = 4.7447 + log ( 0.02 / 0.18 ) =3.79 [H+ ]= 1.62 * 10 ^ -4M When 10 mL of 0.01 M HCl is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases. No.of mols of salt = 0.02 mols - 0.0001 mol = 0.0199 mols No.of mols of Acid = 0.18 mol + 0.0001 mol = 0.1801 mol pH = 4.7447 + log ( 0.0199 / 0.1801 ) =3.78 [H+]= 1.6*10 ^ -4M d) pH = 4.7447 + log ( 0.18 / 0.02 ) =5.69 [H+]= 2.0 * 10 ^-6 M No.of mols of salt = 0.18 mols - 0.0001 mol = 0.1799 mols No.of mols of Acid = 0.02 mol + 0.0001 mol = 0.0201 mol pH= 4.7447 + log ( 0.1799 / 0.0201 ) = 5.69 [H+]= 2.0 * 10 ^ -6 M c) pH = 4.7447 + log ( 0.02 / 0.18 ) =3.79 [H+ ]= 1.62 * 10 ^ -4M When 10 mL of 0.01 M HCl is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases. No.of mols of salt = 0.02 mols - 0.0001 mol = 0.0199 mols No.of mols of Acid = 0.18 mol + 0.0001 mol = 0.1801 mol pH = 4.7447 + log ( 0.0199 / 0.1801 ) =3.78 [H+]= 1.6*10 ^ -4M d) pH = 4.7447 + log ( 0.18 / 0.02 ) =5.69 [H+]= 2.0 * 10 ^-6 M No.of mols of salt = 0.18 mols - 0.0001 mol = 0.1799 mols No.of mols of Acid = 0.02 mol + 0.0001 mol = 0.0201 mol pH= 4.7447 + log ( 0.1799 / 0.0201 ) = 5.69 [H+]= 2.0 * 10 ^ -6 M =5.69 [H+]= 2.0 * 10 ^-6 M No.of mols of salt = 0.18 mols - 0.0001 mol = 0.1799 mols No.of mols of Acid = 0.02 mol + 0.0001 mol = 0.0201 mol pH= 4.7447 + log ( 0.1799 / 0.0201 ) = 5.69 [H+]= 2.0 * 10 ^ -6 M No.of mols of salt = 0.18 mols - 0.0001 mol = 0.1799 mols No.of mols of Acid = 0.02 mol + 0.0001 mol = 0.0201 mol pH= 4.7447 + log ( 0.1799 / 0.0201 ) = 5.69 [H+]= 2.0 * 10 ^ -6 MRelated Questions
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