d
One mole of \(NaOH\) is completely neutralised by one mole of \(HCL\).

Hence, \(0.01\) mole of \(NaOH\) will be completely neutralised by \(0.01\) mole of \(HCL\).

\(\Rightarrow\; NaOH\) left unneutralised \(=0.1-0.01=0.09\) mol As equal volumes of two solutions are mixed,

\([\mathrm{OH}]^{-}=\frac{0.09}{2}=0.045 \mathrm{M} \)

\( \Rightarrow \mathrm{pOH}=-\log (0.045)=1.35 \)

\(\therefore \mathrm{pH}=14-1.35=12.65\)