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 2xx10^(-1)`
C. `1.0`
D. `2.0`

Answer

C

Explanation

50 mL of 0.4 HCl =`0.4/1000xx50`
=0.02 g mol
50 mL of 0.2 M NaOH =`0.2/1000xx50`
=0.01 g mol
0.01 g mol of NaOH will neutralise 0.01 g mol of HCl
HCl left unneutralised = 0.01 g mol.
Vol. of solution =50+50=100 mL
[HCl]=`0.01/100xx1000`=0.1 M
`[H^+]` =0.1 M
pH =-log (0.1)=1.0