The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` is

Options

A. `-log 2`
B. `-log 0.2`
C. `1.0`
D. `2.0`

Answer

C

Explanation

`M_(1)=[HCl]=0.4 mol L^(-1)`
`[ :' 1 NHCl=1 M HCl]`
`M_(2)=[NaOH]=0.2 mol L^(-1)`
`[ :' 1 N NaOH =1 M NaOH]`
`M_(1)V_(1)=(0.4 xx 50)m mol =20 m mol HCl`
`M_(2)V_(2)=(0.2 xx50) m=10 m mol NaOH`
1 mol NaOH `=1` mol HCl
`:. 10m` mol NaOH `-= 10`m mol HCl (NaOH is limiting reactant )
The resultant solution contains 10m mol of HCl in 100mL `(50+50)` of solution.
`[H_(3)O^(+)]=[HCl]=(10xx1000)/(1000xx100)`
`=0.1 M =10^(-1)M`
`pH-log[H_(3)O^(+)]=-log 10^(-1)=1`