The structure of aspartic acid is HO_2CCH_2CHNH_2CO_2H; pK_a1 = 3.86 and pK_a2 =
ID: 510173 • Letter: T
Question
The structure of aspartic acid is HO_2CCH_2CHNH_2CO_2H; pK_a1 = 3.86 and pK_a2 = 9.82 for aspartic acid. Consider an aqueous solution prepared by dissolving 0.0400 g of aspartic acid in enough water to give 50.0 mL of solution. (a) i. What volume of 0.0500 M NaOH must be added to remove the first acidic hydrogen from each aspartic acid molecule (i.e., to reach the first equivalence point)? ii. What is the pH of the resultant solution at the first equivalence point? iii. The indicator phenolphthalein is colorless when added to a pH = 8 solution and red in a pH = 9.5 solution. Would phenolphthalein be a good indicator to signal that the first equivalence point had been reached and addition of titrant should stop? Justify your choice. (b) i. What volume of 0.0500 M NaOH must be added to remove both acidic hydrogens from each aspartic acid molecule (i.e., to reach the second equivalence point)? ii. What is the pH of the resultant solution at the second equivalence point? iii. Would phenolphthalein be a good indicator to signal the second equivalence point?Explanation / Answer
a)
i) No. moles of aspartic acid = n = wt/ molWt = 0.04 g/ 133 = 0.0003 moles
Therefore, we need 0.0003 moles of NaOH needed to reach the first equivalence point.
we have 0.05 M of NaOH. we need to find out how much volume required to get 0.0003 moles
Volume = no. moles / Molarity (moles/L) = 0.0003/0.05 = 0.006 L
Therefore, we need 6 mL of 0.05 M NaOH required to reach the first equivalence point.
ii) pH at the first equivalence point = (pKa1 + pKa2) /2 = (3.86 + 9.82) /2 = 6.84
pH at the first equivalance point = 6.84
iii) Since, the phenolphthalein is colour less below pH 8, and there is no color change in the first equivalance point, it is not a good indicator to detect the first equivalance point.
b)
i)To reach second equivalance point we need to add two times of no.moles of NaOH.
therefore, no. moles of NaOH = 2 X 0.0003 moles = 0.0006 moles
We need to calculate the volume ;
Volume = no. moles / Molarity (moles/L) = 0.0006/0.05 = 0.012 L = 12 mL
Therefore, we need 12 mL of 0.05 M NaOH required to reach the second equivalence point.
ii) At the second equivalance point all the acid [HA-] is converted into [A2-]
[A2-] + H2O <---> HA- + OH-
kb = [HA-][OH-]/[A2-]
[HA-]=[OH-]
[A2-] = 0.0003/0.062L - [OH-] = 0.0048 M - [OH-]= 0.0048 M (since, [OH-] is very small )
pKa2 = 9.82 ; Ka2 = 1.51 x 10-10
Kb = Kw/Ka = 10-14/1.51 x 10-10 = 6.6 x 10-5
6.6 x 10-5 = [OH-]^2/0.0048 M
[OH-]^2 = 31.68 x 10-8
[OH-] = 5.63 x 10-4
pOH = 3.25
pH = 14 - 3.25 = 10.75
The pH at the second equivalance point = 10.75
iii) Again, we can't use phenolphthalein indicator to detect the second equivalance point because after pH 9.5,, the indicator will be red and we can't detect the second equivalance point at 10.75. The indiactor will change colour before it reaches thesecond equivalance point. So, it is not a good indicator.
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