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 0.081M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281?

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 0.081M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281?

aaronq · Answer

equilibrium expression for a general acid dissociation: 

$K_A=\dfrac{[H^+][A^-]}{[HA]}$

You're given pH, we can abstract the concentration of hydronium ions:

$pH=-log[H^+]\rightarrow [H^+]=10^{-pH}=[A^-]$

$K_A=\dfrac{{(10^{-pH}})^2}{(0.081~M-10^{-pH})}$


$pK_A=-log[\dfrac{{(10^{-pH}})^2}{(0.081~M-10^{-pH})}]$