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 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 0.077 M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281? Express your answer numerically.

Answer 1

Step-1

Find out [ H⁺ ] of the weak monoprotic acid using the relation

pH = -log [ H⁺ ]

2.7 = - log [ H⁺ ]

[ H⁺ ] = Antilog (-2.7 )

------ = 1.9953 x 10^-3

Since , X-281 is a weak monoprotic acid , its equiliibrium constant ( K_a ) can be calculated using Ostwald's dilution law - relation as-

K_a = c²   where , c represents concentrtion of acid in moles per litre (given as 0.077M ) and represents degree of dissociation of the acid

So, K_a = 0.077 x { 1.9953 x 10^-2 / 0.077 }²  

------------- = 5.1704 x 10^-3  

Now apply the relation pK_a   = -log K_a  to get the pK_a value of X-281

Thus ,pK_a   = -log ( 5.1704 x 10^-3 )

-------------- = 2.2865

or,---------- = 2.29