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 $A\times {10}^B$.
$A+B$ is ............

Answer 1

First obtain the stepwise formation constants from logK 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.
$K=K_1\times K_2\times K_3={10}^{4.34}\times {10}^{3.31}\times {10}^{2.05}={10}^{9.7}=5\times {10}^9$, this is equal to $A\times {10}^B$.

Hence, $A=5$ and $B=9$

$A+B=9+5=14$