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

Correct Answer - C
`N_(1)V_(1)-N_(2)V_(2)=NV`
`0.1xx1-0.01xx1=Nxx2`
`[OH^(-)]=N_(R)=(0.09)/(2)=0.045 N`
`pOH= -log (0.045) =1.35`
`:. pH=14- pOH=14-1.35=12.65`
`:. S=10^(-3) M`

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

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