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In an amino acid, the carboxylic group ionises at $\mathrm{pH}=2.56\left(\mathrm{pK}_{\mathrm{a}_1}\right)$ and ammonium ion ionises at $\mathrm{pH}=9.38\left(\mathrm{pK}_{\mathrm{a}_2}\right)$. The isoelectric point of the amino acid is at
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The correct answer is:
5.97
$\mathrm{p} K_{a 1}=256, \mathrm{p} K_{a_2}=9.38$
At isoelectric point
$$
(\mathrm{pH})=\frac{\mathrm{p} K_{a_1}+\mathrm{p} K_{a_2}}{2}=\frac{2.56+9.38}{2}=5.97
$$
At isoelectric point
$$
(\mathrm{pH})=\frac{\mathrm{p} K_{a_1}+\mathrm{p} K_{a_2}}{2}=\frac{2.56+9.38}{2}=5.97
$$
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