Question

For the ionization of a weak acid, HA:
a) give the expression for Ka. b) after taking the log 10 of both sides ofthe quation in "a" sp;ve fpr pH. c) under what conditions would pH be equal to pK?

Answer 1

This is just the derivation of the Henderson-Hasselbalch equation.
I'll get you started.
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Now do what b says. Take the log of both sides.
log Ka = log(H^+)(A^-)/(HA) or
log Ka = log[(A^-)/(HA)] + log(H^+)
Multiply through by -1
-log Ka = - log([A^-)/(HA)] -log(H^+)
But note that -log Ka = pKa and -log(H^+) = pH. Now you finish. You should end up with
pH = pKa + log[(A^-)/(HA)]
Post your work if you get stuck.
Obviously, pH = pKa when [(A^-)/(HA)] = 1

Answer 2

thank you!