A solution of weak acid HA is being titrated with 0.1M $NaOH$  solution. End point is reached on addition of x mL of  $NaOH$. To the above solution,   $\frac{x}{2}mL$ of 0.10M  HCl is added further then the pH of resulting solution was found to be 5. The  pH at the  $NaOH$ end point if initial concentration of acid was 0.10M is ___________(round off to nearest integer) 

Our first objective here is to find $K_a$  of the weak acid. From initial concentration of acid, base and end point volume of $NaOH$ , is evident that initial volume of acid was also $xmL$ , At  $NaOH$ end point, mmol of  $NaA$ salt  $=0.10x.$ After adding  $HCl$, following neutralization will occur:
        $HCl+NaA\to NaCl+HA$       
Initial: $\frac{0.10x}{2}$ $0.01x$ B $0$  $0$                    
Final:   $0$           $\frac{0.10x}{2}$  $\frac{0.01x}{2}$  $\frac{0.01x}{2}$                  
The above solution is a buffer with same concentration of HA and its conjugate base. Applying Henderson’s equation
   $pH=p{K}_a+\log \frac{\left[A^{-1}\right]}{\left[HA\right]}=p{K}_a=5\Rightarrow K_a={10}^{-5}$
Now, at the $NaOH$  and point, it is solution of NaA where   hydrolyses as:
 $$\begin{gathered} A^{-}+H_{2}O\underset{ }{\rightleftharpoons }HA+O{H}^{-} \\ {K}_h=\frac{k_w}{k_a}={10}^{-9} \\ \Rightarrow \left[O{H}^{-}\right]=\sqrt{K_{h}C}=\sqrt{{10}^{-9}\times 0.05}={10}^{-5} \end{gathered}$$
 $\Rightarrow P^{OH}=5.15$  and  $P^H=8.85$