Question

A 1.25 × 10-4 m solution of the anti-inflammatory drug naproxen has a ph = 4.2. the ka and pka of naproxen are ________ and ________.

Answer 1

Answer:

$Ka=6.43x10^{-5}$

$pKa=4.19$

Explanation:

Hello,

In this case, the pH provides the concentration of hydrogen in its dissociation:

$Naproxen\rightleftharpoons H^++Naproxen^-$

Thus, such concentration is:

$[H^+]=10^{-4.2}=6.31x10^{-5}M$

Thus, the Ka, by means of the dissociation law of mass action:

$Ka=\frac{(6.31x10^{-5})^2}{1.25x10^{-4}-6.31x10^{-5}}=6.43x10^{-5}$

Hence the pKa is:

$pKa=-log(Ka)=-log(6.43x10^{-5})=4.19$

Best regards.

Answer 2

Naproxen is known to be a weak acid. In order to calculate its ka and pka, use the equation of getting the ph of weak acid which is ph= -1/2 log [(Ka)(Mwa)]. The Ka value is 3.18x10^-5. The pKa can be obtained through pKa = - log Ka. The pKa is 4.5.