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

Question

Part A 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 K_a value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries. To find the pK_a of X-281, you prepare a 0.076 test solution of X-281 at 25.0 degree C. The pH of the solution is determined to be 2.40. What is the pK_a of X-281? Express your answer numerically.

Explanation / Answer

It is given that the heart medicine is a monoprotic weak acid.

1) Write the dissociation equation for the acid:

HA H+ + A¯

2) Write the equilibrium expression:

Ka = ( [H+] [A¯] ) / [HA]

3) Our task now is to 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¯2.4 =3.9 x 10^-3 M

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

[A¯] = 3.9 x 10^-3 M

c) the final value, [HA] is given in the problem. It is 0.076M

1) Ka = [3.9 x 10^-3 M x 3.9 x 10^-3 M ] /0.076 M

        = 200.13 x 10^-6 M

2) pKa = -log ( 200.13 x 10^-6 M)

          = 3.69

          = 3.7 (to two significant figures)

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