Question

For a certain diprotic acid, the pH at one half the volume to the first equivalence point is 4.15 and the pH at one half the volume between equivalence points is 7.22. Given this information, what are the values for Ka1 and Ka2

Answer 1

Answer:

The values for

$Ka_{1} = 10^{-4.15}= 7.07 X 10^{-5}$ & $Ka_{2}=10^{-7.22}= 6.02 X10^{-8}$

Explanation:

               Dissociation of diprotic acid is as follows

                      $H_{2}A(aq) = H^{+}(aq) + HA^{-} (aq)$   ----------------(i)

                    $HA^{-}(aq)=H^{+}(aq) + A^{2-}(aq)$      ---------------(ii)

At the first equivalence point 50% of the acidic proton is lost , then it has the other 50% of its conjugate base.

  From equation (i) we can write

               $Ka_{1}= \frac{[H^{+} ][HA^{-} ]}{[H_{2}A ]}$

               $pH_{1} = -logKa_{1} + log\frac{[Base]}{[Acid]}$

                $4.15 = -logKa_{1}+log\frac{0.5}{0.5}$            

              ∵ At equivalence point [Conjugate base = Acid]  

                $Ka_{1} = 10^{-4.15}= 7.07 X 10^{-5}$

   From equation (ii) we can also write

               $pH_{2}=-log Ka_{2} + log\frac{[Base]}{[Acid]} \\pH_{2}=-log Ka_{2} + log\frac{[A^{2-} ]}{[HA^{-} ]}\\\\7.22 = - logKa_{2} +log(1)\\\\Ka_{2}=10^{-7.22}= 6.02 X10^{-8}$