I don\'t understand #30 thanks in advance for the help 8 13 Inge Department For

Question

I don't understand #30 thanks in advance for the help

8 13 Inge Department For questions 2831 consider the amino acid valine with the chemical CoHoNHcooH At neutral pH valine an be thoughtofas an a species because it can donate the proton mpboteric fromtts so00H group and It can accept a protononits NHzgroup. At low pH Valint can act as a dipretie add (which means that it can donate 2 consecutive protons)and at high pH itcan act as diprotic base (which means it can accept consecutive protons) The titration to the right corresponds to the titration of diprotic valine with 0.1 M Na0H. 28)Use the titration curve to estimate the two pKa values of valine. ai 14 and 3.4 b. 2.4 and 9.7 24 and 60 d, 60 and 11 ies 97 and 11 29)Which of the following represents valime in its most acidic form (as a diprotic acid)? NH 30)if you have a solution of valine at pH 90, approximately how many moles of 0.t M NaoH can you add before you coded the buffer capacity 00001 moles b. 00003 moles c, 00005 moles 0.0007 moles 00009 moles 31)At pH 18 valine exists in its most agidic, diprotik form. How many mla or02 M NaoH will you need to add to a solution of valine at pH 18 to fulo deprotonate the valune? S ml. 10 mL d, 40 ml.

Explanation / Answer

Valine a diprotic acid titration with 0.1 M NaOH

28) The pKa is half way pH to the equivalence point.

From the plot the pKa of valine would be,

b) 2.4 and 9.7

29) Most acidic form of the amino acid will have a protonated amine group along with undisscoiated carboxylic acid group,

a.

30) For the titration,

let x moles of base is added,

using hendersen-hasselbalck equation,

pH = pKa + log(base/acid)

9 = 9.7 + log(x/(0.1 M x 20 ml - x)

10 - 5x = x

x = 1.7 mmol

acid left in solution = 2 - 1.7 = 0.3 mmol = 0.0003 mole

So the moles of NaOH to be added would be,

b. 0.0003 moles.

31) Looking at the graph,

A fully protonated valine would need 2 moles of NaOH for complete neutralization

Thus,

moles of base added when 0.1 M NaOH used = 20 ml

Volume of base needed when 0.2 M NaOH was used = 0.1 M x 20 ml/0.2 M = 10 ml

b. 10 ml

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.