50 ml of 0.05 M `HNO_3`is added to 50 ml of `0.025 M KOH`. Calculate the pH of the resultant solution.

Explanation

Number of moles of `HNO_3 =0.05xx50xx10^-3`
`=2.5xx10^-3`
Number of moles of KOH `=0.025 xx50xx10^(-3)`
`=1.25xx10^-3`
number of moles of `HNO_3` after mixing
`=2.5xx10^(-3)-1.5xx10^(-3)`
`=1.25xx10^(-3)`
`therefore` concentration of `HNO_3=("Number of moles of"HNO_3)/("Volume is litre")`
After mixing, total volume `= 1000 ml =100xx10^(-3)L`
`therefore [H^+]=(1.25xx10^(-3)"moles")/(100xx10^(-3)L)`
`=1.25xx10^2` moles `L^(-1)`
`pH=-log[H^+]`
`pH=-log(1.25xx10^(-2))=2-0.0969`
`=1.9031`.