John prepares a sample of 100mL of 0.1M leucylhistidine at pH 6.48. You then tel
ID: 53090 • Letter: J
Question
John prepares a sample of 100mL of 0.1M leucylhistidine at pH 6.48. You then tell him that you actually wanted pH 7.00. How many milliliters of 1.0 M NaOH does he have to add? (Assume that the pKa 's correspond to the values at the separate amino acids)
Leucine 2.33 9.74
Histidine 1.80, 6.00, 9.33
Please give a detailed description and clear work including units at every step, and also explain how and why you choose which pKa to use in the Henderson Hassel Equation. Thank you
A. 0.60 mL
B 1.59 mL
C. 2.83 mL
D) 2.92 mL
E) 5.18 mL
Explanation / Answer
pH = pKa + log [ base / acid]
6.48 = 6.00 + log[base/acid]
[base] / [acid] = 3.02
[base] = [acid] * 3.02
millimoles of buffer = 100 * 0.1 = 10 millimol
millimol of base + millimol of acid = 10
millimol of acid * 4.02 = 10
nillimol of acid = 10/4.02 = 2.488
millimol of base = 7.52 millimol
If pH = 7
7 = 6 + log[base/acid]
[base] / [acid] = 10
millimol of acid + millimol of base = 10
11 * millimol of acid = 10
millimol of acid = 0.909
millimol of base = 9.909
Change in millimol of base = millimol of NaOH added = 9.909 - 7.52 = 2.389 millimol
Volume of NaOH = millimol / Moalrity = 2.389 mL
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