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 0.089 M test solution of X-281 at 25.0 C. The pH of the solution is determined to be 3.00. What is the pKa of X-281?

Explanation / Answer

Since X-281 is a weak monoprotic acid ,let us represented it as HA.

1) Write the dissociation equation for the acid:

HA H+ + A¯

2) Write the equilibrium expression:

Ka = ( [H+] [A¯] ) / [HA]

3) Determine the three concentrations on the right-hand side of the equilibrium expression since the Ka is our unknown.

a) We will use the pH to calculate the [H+]. We know pH = -log [H+], therefore [H+] = 10¯pH

[H+] = 10¯3 = 1 x 10^-3 M

b) From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [A¯]. Therefore:

[A¯] = 1 x 10¯3M

c) the final value, [HA] is given in the problem. In the example being discussed, 0.089M is the value we want. Some teachers will use 0.089 M while others would say to subtract the 1 x10^-3 M value from 0.089 M. Let's do both.

c1) Ka = [(1 x 10¯3) (1 x 10¯3] / 0.089

Ka = 1.12 x 10¯5

c2) Ka = [(1 x 10¯3) (1 x 10¯3)] / (0.089 - 1 x 10¯3)

Ka = 1.12 x 10¯5

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