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 KaKaK_a value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries. To find the pKapKapKa of X-281, you prepare a 0.079 MM test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKapKapKa of X-281

Answer 1

Answer:

The $pK_a$ of the X-281 medicne is 4.29.

Explanation:

Concentration of medicine in solution  = $[HA]=c=0.079 M$

$HA\rightleftharpoons A^+H^+$

Initially

c       0       0

At equilibrium

(c-x)    x     x

The expression of an dissociation constant $K_A$ is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(c-x)}$..[1]

The pH of the solution = 2.70

$pH=-\log[H^+]$

$2.70=-\log[H^+]=-\log[x]$

$x=10^{-2.70}=0.001995 M$..[2]

Using [2] in [1]

$K_a=\frac{0.001995 M\times 0.001995 M}{(0.079-0.001995 M)}$

$K_a=5.170\times 10^{-5}$

The $pK_a$ of the medicine :

$pK_a=-\log[K_a]=-\log[5.170\times 10^{-5}]=4.29$

The $pK_a$ of the X-281 medicne is 4.29.