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 0.086 M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281?

Explanation / Answer

Solution

Given data

Concentration of the X-281 = 0.086 M

pH= 2.70

pka = ?

from the given pH value we can calculate the concentration of the hydronium ions that are produced by the X-281 after the dissociation

lets assume X-281 = HA

now lets write the its dissociation equation

HA + H2O ----- > H3O+ + A^-

Now lets calculate the [H3O+] using the pH

pH= -log [H3O+]

therefore

[H3O+] = antilog [ -pH]

             = antilog [ - 2.70]

             = 0.001995 M

The concentration of the H3O+ is same as concentration of the A^- = 0.001995 M

Now lets write the ka equation

Ka= [H3O+][A^-] /[HA]

Lets put the values in the formula

Ka = [0.001995] [0.001995] / [ 0.086]

Ka= 4.63*10^-5

Now using the ka value lets calculate the pka of the X-281

Pka = -log [ka]

Pka = - log [ 4.63*10^-5]

Pka = 4.33

Therefore the pka of the X-281 = 4.33

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.