${Pk}_{a} and {Pk}_{b} values of isoleucine is 2.36 and 9.68 respectively . Then P_{I} = ?$

2.36

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9.68

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6.02

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3.66

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Solution

The correct option is C 6.02
$P_{I} = \frac{1}{2} ({Pk}_{a} + {Pk}_{b})$

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{Pk}_{a} and {Pk}_{b} values of isoleucine is 2.36 and 9.68 respectively . Then P_{I} = ?

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