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

Options

A. 7
B. 1.04
C. 12.65
D. 2

Answer

C

Explanation

Let volume of each solution is V.
Acid=Eq. of HCl=0.01`xxV`, Base=Equivalent of NaOH=0.1`xxV`
Net equivalent=0.1V-0.01V
`=V[0.1-0.01]`
=0.09V
`thereforeNxx2V=0.9V` (basic)
`[OH]^(Theta)N=(0.09)/(2)=4.5xx10^(-2)`
`pOH=2-log4.5`
`=2-0.65`
=1.35 ltbr `thereforepH=14-1.35=12.65`