The following chemical equations shows the stepuise formation constant expression $(k_1$ to $k_3)$ of each $ox$ replaces to water

$Fe+C_2{O}_4^{2-}[Fe(C_2{O}_4){]}^{+}$

$\{\begin{array}{c}\text{:}[Fe(C_2{O}_4){]}^{+}+C_2{O}_4^2[Fe(CS{O}_4{)}_2{]}^{-}\end{array}$

$\{\begin{array}{c}\text{:}[Fe(C_2{O}_4{)}_2{]}^{-}+Cr{O}_4^2[Fe(C_2{O}_4{)}_3{]}^{3-}\end{array}$

$K_1=\frac{[Fe(Ox)f]}{[F{e}^{3+}][0x]},K_2=\frac{[Fe(ox{)}_2^{-}]}{[Fe(ox{)}^{}][0x]},K_3=\frac{[Fe(ox{)}_3^{3-}]}{[Fe(Ox{)}_2^{-}][0x]}$

a$)$ Write the overall formation constant experssion or $($:${}_3\}$ demonstrating that the $K_F$ is given by the product of the three stepwise constants.