points Gvsuchem 1162 14h 3PO01 Polyprotic Acids: Amino Acids In its fully proton
ID: 499627 • Letter: P
Question
points Gvsuchem 1162 14h 3PO01 Polyprotic Acids: Amino Acids In its fully protonated form an amino acid acts like any other polyprotic acid. ir we titrate this polyprotic acid, we could determine the pKa for each of the potentially that amino acid. Consider the amino acid tyrosine shown below in both its neutral form and its fully protonated form as a Lewis structure and as an Site 2 Neutral form of tyrosine Lewis dot structure Fully protonated form of tyrosine Neutral of Tynesine form present at low pH Lewis dot structure Here it can act as a a) the fully protonated form, tyrosine has three acidic sites. Therefore, tyrosine will have three equivalence Ir you start with a 10.0 of In required to completely remove one or the acidic protons reach an 0.10 M tyrosine, and you titrate it with 0.10 M NaoH, how many mL of NaoH are equivalence point)?Explanation / Answer
Tyrosin has three ionizable protons; denote tyrosin as H3A. Since there are 3 ionizable protons, there will be three distinct point on the pH vs volume curve corresponding to the three protons.
a) We start with 10.0 mL of 0.10 M tyrosin.
Moles of tyrosin taken = (volume of tyrosin taken in L)*(concentration of tyrosin taken in mol/L) = (10 mL)*(1 L/1000 mL)*(0.10 mol/L) = 0.0010 mole.
Write down the balanced chemical equation for the complete neutralization of tyrosin with NaOH.
H3A + 3 NaOH -----> Na3A + 3 H2O
As per the balanced chemical equation, we have 1 mole tyrosin = 3moles NaOH.
Therefore, 0.0010 mole tyrosin = (0.0010 mole tyrosin)*(3 moles NaOH/1 mole tyrosin) = 0.0030 mole tyrosin.
Concentration of NaOH used = 0.10 M
Therefore volume of NaOH required = (moles of NaOH required)/(concentration of NaOH in mol/L) = (0.0030 mole)/(0.10 mol/L) = 0.030 L = (0.030 L)*(1000 mL/1 L) = 30 mL (ans).
b) Since H3A is a triprotic acid, the conjugate bases are H2A-, HA2- and A3-.
A total of 30 mL of NaOH is required; therefore, 10 mL of NaOH is required to reach the first equivalence point (here the predominant species is H2A-) and pH = pKa1.
Look at the titration curve and find that the pH corresponding to 10.0 mL of NaOH is roughly 3.5. Therefore,
pKa1 = 3.5 (ans)
A second 10.0 mL is required to reach the second equivalence point and we have the relation
pH = ½*(pKa1 + pKa2)
The pH corresponding to 20.0 mL is roughly 7.0 and pKa1 = 3.5. Therefore,
7.0 = ½*(3.5 + pKa2)
===> 14.0 = 3.5 + pKa2
===> pKa2 = 14.0 – 3.5 = 10.5 (ans)
Another 10.0 mL is required to reach the third equivalence point and the pH (from the curve) corresponding to the third equivalence point is 11.5. At the third equivalence point, we must have
pH =1/2*(pKa2 + pKa3)
===> 11.5 = ½*(10.5 + pKa3)
===> 23.0 = 10.5 + pKa3
===> pKa3 = 12.5 (ans).
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