BIO370 Homework Set # 1: Buffers C-terminus ofamino acid i C-terminus in peptide
ID: 575389 • Letter: B
Question
BIO370 Homework Set # 1: Buffers C-terminus ofamino acid i C-terminus in peptide Spring 2018 Phosphoric acid Sodium dihydrogen phosphate 7.2 Disodium hydrogen phosphate 12.3 Asp Citric acid Sodium dihydrogen citrate Disodium hydrogen citrate Glycylglycine MOPS MES HEPES Tris 3.1 4.7 3.7 4.25 N-terminus of amino ac N-terminus in peptide Tyr Lys Art His 5.4 8.4 7.2 6.2 7.6 8.3 8.2 10 10 10.5 12.5 1. What is the pH of a solution with a potassium acetate cacetic acid molar ratio of 5:1 (pK-4.7)? a. If the total concentration of acetate (that is, of acetic acid + acetate) is 100 m each component in the solution? b. You add NaOH to the above solution to a final NaOH concentration of 30 mM. What is the pH nowr out the equilibrium. Remember ICE from Gen Chem? Use that here.)Explanation / Answer
1.
Using Henderson-Hesselbalach equation
pH = pKa + log { [salt] / [acid] }
or, pH = pKa + log { [CH3COOK] / [CH3COOH] }
Given, pKa = 4.7 and
[CH3COOK] : [CH3COOH] = 5:1
or [CH3COOK] / [CH3COOH] = 5/1 = 5
Now, substituting the values, we get;
pH = pKa + log { [CH3COOK] / [CH3COOH] }
= 4.7 + log (5)
= 4.7 + 0.70
= 5.4
(a)
We know that
[CH3COOK] / [CH3COOH] = 5/1 = 5
[CH3COOK] = 5 [CH3COOH]
Given;
[CH3COOK] + [CH3COOH] = 100 mM = 0.1 M
So,
[CH3COOK] + [CH3COOH] = 0.1
or, 5 [CH3COOH] + [CH3COOH] = 0.1
or, 6 [CH3COOH] = 0.1
or, [CH3COOH] = 0.1 / 6
or, [CH3COOH] = 0.017 M
Hence, [CH3COOK] = 5 [CH3COOH]
= 5 x 0.017 M
= 0.085 M
(b)
Considering 1 liter solution
NaOH added = 30 mM = 0.03 M = 0.03 mole
[CH3COOH] = 0.017 M = 0.017 moles
So, 0.017 moles of CH3COOH will be neutralized with 0.017 moles of NaOH.
Moles of NaOH left = 0.03 moles – 0.017 moles = 0.013 moles = 0.013 M
So, pOH = - log [OH-]
or, pOH = - log (0.013)
or, pOH = 1.89
pH = 14 – pOH
= 14 – 1.89
= 12.11
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