What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

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.

NaOH left unneutralised = 0.1 - 0.01 mol

                                                   = 0.09 mol

As equal volumes of two solutions are mixed,

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

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

$\therefore \mathrm{pH}=14-1.35=12.65$