A new potential heart medicine, code-named X-281, is being tested by a pharmaceu

Question

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries. To find the pKa of X-281, you prepare a 0.079 M test solution of X-281 at 25.0 C. The pH of the solution is determined to be 2.70. What is the pKa of X-281?

Explanation / Answer

pH of the 0.079 M solution of test solution of X-281 is 2.70.

From pH we can find out H+ ion concentration i.e. [H+]

pH = -log[H+] ..............(since, [H+] = [H3O+])

2.70 = -log[H+]

log[H+] = -2.70

[H+] = 10-2.70

[H+] = 2.00 x 10-3 M/L (apprx.)

Let HA be that acid whose ionization equation is given as,

HA <-------------> H+ (aq.) + A- (aq.)

Dissociation constant Ka is given as,

Ka = [H+][A-]/[HA]

We have initially,

[HA] = 0.079 M

on ionization at equilibrium we have,

[H+] = [A-] = 2.00 x 10-3 M

And hence [HA] = 0.079 - 2.00 x 10-3 = 0.077 M

Let us put these values at equilibrium in expression for Ka,

Ka = (2.00x10-3)(2.00x10-3)/(0.077)

Ka = 4.00x10-6/0.077

Ka = 5.19 x10-5.

Then we define pKa as,

pKa = -log(Ka)

pKa = -log(5.19 x10-5)

pKa = 4.28

Hence pKa of X-281 drug is 4.28.

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