Calculate the pH of a solution which contains 100 ml of 0.1 H HCl and 9.9 ml of 1.0 M NaOH.

Explanation

Here 100 ml of 0.1 M HCl
meq `=100xx0.1=10`
9.9 ml of 1.0 M NaOH
meq`=9.9xx9.9` meq
meq of acid left k`=10-9.9=0.1`
Normality of acid `=("meq")/("Total Vol")`
`=0.1/109.9=9xx10^(-4)`
`pH=3.0416`