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) = 0.05 xx 50 xx 10^(-3)`
`= 2.5 xx 10^(-3)`
Number of moles of KOH = ` 0.025 xx 50 xx 10^(-3)`
` 1.25 xx 10^(-3)`
Number of moles of ` HNO_(3)` after mixing
2.5 xx 10^(-3) - 1.5 xx 10^(-3)`
`1.25 xx 10^(-3)`
Concetration of ` HNO_(3)`
` = ( " Number of moles of ` NHO_(3)) /( " Volume of littre") `
After mixing , total volume =` 100 ml = 100 xx 10^(-3) L `
` [ H^(+)] = ( 1.25 xx 10^(-3) " moles")/( 100 xx 10^(-3) L)`
` 1.25 xx 10^(-2) " moles " L^(-1)`
pH= `log [ H^(+)]`
` pH= - log ( 1.25 xx 10^(-2)) = 2- 0.0969`
` = 1.9031 `