`20 cm^(3)` of `xM` solution of `HCI` is exactly neutralised by `40cm^(3)` of `0.05 M NaOH` solutions, the `pH` of `HCI` solution is

Options

A. 1
B. 2
C. 1.5
D. 2.5

Answer

A

Explanation

Moles of NaOH `=0.05xx40xx10^(-3)`
Moles of HCl `=20xx 10^(-3)xx x`
`0.05 xx 40 xx 10^(-3)=20 xx x xx 10^(-3)`
or `x=(0.05xx40)/(20)=0.1 ` or `1xx10^(-1)`
`:. [HCl]=[H_(3)O^(+)]=1xx10^(-1)`
`:. pH =-log (1xx10^(-1))=1`