The pl (isoelectric point) of aspartic acid is:
JEE Main Online Paper (Held On 09 April 2016)
A) 5.74
B) 3.65
C) 2.77
D) 1.88
Correct Answer: C
Solution :
In given reaction sequence \[Pl=\frac{p{{K}_{1}}+p{{K}_{R}}}{2}\] \[=\frac{1.88+3.65}{2}=2.77\]
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