Answer to Question #310713 in General Chemistry for donya

Question

a)

An increase in the ionisation of a molecule in a GI environment will usually

increase its solubility. Valproic acid is an oral dosage form used in epilepsy and

has a pKa of 4.8. With the stomach pH at 2.0 and the ileum at pH 6.8, which area

of the GIT will Valproic acid be more soluble.

b)

You are required to manufacture 1L of 50mM pH 7.0 phosphate buffer that could be

used for an injectable formulation of valproic acid. The Ka of dihydrogen phosphate is

6.3x10

-8

and the RMM of potassium dihydrogen phosphate and dipotassium

hydrogen phosphate are 136 and 174 respectively.

I.

Calculate the pKa of dihydrogen phosphate.

II.

What is the ratio between the conjugate base and the acid to be used?

III.

Calculate the mass of anhydrous potassium dihydrogen phosphate and

dipotassium hydrogen phosphate with RMM of 136 and 174, and

concentrations of 30.5 and 19.5 mmol respectively.

Answer 1

a) pH = pK_a + log₁₀{[A^-]/[HA]}

Stomach (pH = 2.0)

2.0 = 4.8 + log₁₀{[A^-]/[HA]}

-2.8 = log₁₀{[A^-]/[HA]}

2.8 = log₁₀{[HA]/[A^-]}

631x + x = [A]_total

99.84% is protonated and 0.16% is unprotonated

Ileum (pH = 6.8)

6.8 = 4.8 + log₁₀{[A^-]/[HA]}

2.0 = log₁₀{[A^-]/[HA]}

100x + x = [A]_total

99.00% is unprotonated and 1.00% is protonated

In stomach Valproic acid is more soluble

b) I. pK_a = -log₁₀ K_a = -log₁₀ (6.3×10^-8) = 7.2

II. pH = pK_a + log{[A^-]/[HA]}

[HA] + [A^-] = 0.05

7.0 = 7.2 + log{(0.05-x)/x}

x/(0.05 - x) = 1.58

x = 0.0306

[HA] = 0.0306 mol

[A^-] = 0.0194 mol

[A^-]/[HA] = 1.577

III. For molecular weights: KH₂PO₄ = 39 + 97 = 136 amu; K₂HPO₄ = 39×2 + 96 = 174 amu

Thus, 0.0306×136 = 4.161 g of KH₂PO₄ and 0.0194×174 = 3.376 g of K₂HPO₄ should be added.