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 Kavalue 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 0.083M test solution of X-281. The pH of the solution is determined to be 3.00. What is the pKa of X-281?

Please help me solve this. I have looked at other examples but dont understand it. Please show work if possible

Thank you

Explanation / Answer

pKA = -log(Ka)

If weak acid:

HX <-> H+ and X-

The equilibrium expresion:

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

if pH = 3 then

pH = -log(H+)

3 = -log(H+)

10^-3 = [H+]

0.001 = [H+]

From equilibrium

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

Assume [H+] = [A-] = 0.001

And that

[HA] = 0.077 - 0.001 =  0.076

Therefore

Ka = [H+][A-] /[HA] = 0.001*0.001 / (0.076) = 1.31*10^-5

But we need pKa

pKa = -log(Ka) = -log(1.31*10^-5 ) = 4.88

pKa = 4.88

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