Question

Ethylenediamine displaces $H_{2}O$ in the complex $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ in three steps.
$\left[Fe\right(H_{2}O{)}_6{]}^{3+}+3en\to \left[Fe\right(en{)}_3{]}^{3+}+6H_{2}O$
Stepwise formation constants are:
$\log K_1=4.34$
$\log K_2=3.31$
$\log K_3=2.05$
Overall formation constant is $A\times {10}^{2B}$.
$\text{B}$ in question is:

Answer 1

First, obtain the stepwise formation constants from log K values.

$\log K_1=4.34$. Hence, $K_1={10}^{4.34}$
$\log K_2=3.31$. Hence, $K_2={10}^{3.31}$
$\log K_3=2.05$. Hence, $K_3={10}^{2.05}$

Now calculate the overall formation constant.

$K=K_1\times K_2\times K_3$

$={10}^{4.34}\times {10}^{3.31}{10}^{2.05}$

$=5\times {10}^{10}$.

But this is equal to $A\times {10}^{2B}$.

Hence, $B=5$