Example

Tryptophan has two pKa values: 2.4 and 9.4. Calculate the solubility of tryptophan at pH 10 and at pH 2, given that the solubility of the compound in neutral solutions is 2 x 10 2 mol dm 3.

Answer

S0 = 2 x 10 2 mol dm We must use equations

2x 10 That is,

Therefore, S = (5.02 x 10~2) + (2 x 1002) 7.02 x 10 2 mol dm 3.

At pH 10 (using equation 5.24)

That is,

Therefore, S = (7.96 x 10~2) + (2 x 10~2) 9.96 x 10 2 mol dm 3.

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