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In an amino acid, the carboxyl group ionises at $\mathrm{p} K_{a_1}=2.34$ and ammonium ion at $\mathrm{p} K_{a_2}=9.60$. The isoelectric point of the amino acid is at pH
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The correct answer is:
$5.97$
Isoelectric point, $\mathrm{pH}=\frac{\mathrm{p} K_{a 1}+\mathrm{p} K_{\propto}}{2}$
$=\frac{2.34+9.60}{2}=5.97$
$=\frac{2.34+9.60}{2}=5.97$
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