Question

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3. Drug A and Drug B are both weak bases, and K. (Drug A) = 2.6x106 and K. (Drug B) = 9x10 a. Calculate pka (Drug A) and pK, (Drug B). b. Which base is stronger?

Answer 1

3.

a) Kb is the equilibrium constant for the protonation of a base, in equilibrium with its neutral form. pK is the negative logarithmic measure of the equilibrium constant (pKb = -logKb).

pH of a substance is calculated usually in water unless otherwise mentioned. Due to the autolysis of water, the concentration of hydroxide ions and hydronium ions at NTP are equal (1x10^-7 M) which gives the pH = pOH of water to be 7. pKw of water, which is the negative logarithmic measure of the autolysis rate of water is given as pKw = pH + pOH = 14. This allows relation of pH and pOH of an acid or base in their aqueous solutions.

So, here pKa = 14 - pKb.

For Drug A: pKb = -log(2.6x10^-6) = 5.5850 which gives its pKa to be 14 - 5.5850 = 8.415.

Similarly, for Drug B, pKb = 14 - {-log(9x10^-7)} = 14 - 6.0457 = 7.9543.

b) The strength of a base or acid can be compared by considering its rate of protonation and ionization respectively. As elucidated earlier, Kb is the equilibrium constant of a base with its neutral and protonated form with the protonated form being the product. Thus, a base with greater basicity will have a larger concentration of the product, which will give it higher Kb value.

Therefore, of the two given drugs, the one with higher Kb is the stronger base. That is Drug A.