Question

He pka values of a compound with two ionizable groups are pk1 = 4.10 and pk2 between 7 and 10. a biochemist has 10 ml of a 1.0 m solution of this compound at a ph of 8.00. she adds 10.0 ml of 1.00 m hcl, which changes the ph to 3.20. what is pk2?

Answer 1

For the reaction of weak acid:

$HA\rightleftharpoons H^{+}+A^{-}$

The Henderson-hasselbalch equation is:

$pH = pk_a+ log \frac{[A^{-}]}{[HA]}$

The concentration ratio at $pk_1$ is:

$pH = pk_1+ log \frac{[A^{-}]}{[HA]}$

Substituting the values:

$3.2 = 4.1 + log \frac{[A^{-}]}{[HA]}$

$log \frac{[A^{-}]}{[HA]} = - 0.9$

$\frac{[A^{-}]}{[HA]} = 0.126$

Now, for calculating the pk_2 value:

$pH = pk_2+ log \frac{[A^{-}]}{[HA]}$

$8 = pk_2+ log(0.126)$

$pk_2 = 8.9$

Thus, $pk_2 = 8.9$.