What would be the pH of a solution obtained by mixing 100 ml of 0.1 N HCl and 9.9 ml of 1.0 N NaOH solution ?

What would be the pH of a solution obtained by mixing 100 ml of 0.1 N

1.	What would be the pH of a solution obtained by mixing 100 ml of 0.1 N HCl and 9.9 ml of 1.0 N NaOH solution ?
Answer» Solution :100 ml of 0.1 N HCl = `100xx0.1 ` milli eq. = 10 milli eq. 9.9 ml of 1 N NaOH `=9.9xx1 ` milli eq. = 9.9 milli eq. `:.` HCl LEFT unneutralized `=10-9.9= 0.1` milli eq. Volume of solution = 100 + 9.9 = 109.9 ml `:.` Normality pf HCl in RESULTING solution `= (0.1)/(109.9)~~(0.1)/(110)=0.09xx10^(-4)N = 9.09 XX 10^(-4)M` As HCl completely ionizes as `HCl + H_(2)O RARR H_(3)O^(+) + Cl^(-)` `:. [H_(3)O^(+)]=[HCl]=9.09xx10^(-4)M` `pH = - LOG [H_(3)O^(+)]=- log (9.09xx10^(-4))=- [ log 9.09 + log 10^(-4)]= - [0.9546-4]` `=3.0454 = 3.05`