Neutralization constant of HCOOH with a strong base is ${10}^8$ . What is the pH of 0.01M HCOOK solution at ${25}^{°}C$ ?

$HCOOH+NaOH\to HCOONa+H_{2}O$ ; $K_{eq}={10}^8$  
$HCOOH+O{H}^{-}\to HCO{O}^{-}+H_{2}O$ ; $K_{eq}={10}^8$ 
So, for the hydrolysis reaction $[Kh={10}^{-8}]$
$\left[O{H}^{-}\right]=Ch=\sqrt{Kh.C}$
$$\begin{gathered} =\sqrt{{10}^{-8}\times {10}^{-2}} \\ =\sqrt{{10}^{-10}} \\ ={10}^{-5} \\ {P}^{OH}=5 , \\ {P}^H=9 \end{gathered}$$