What will be the resultant pH, when 200 mL of an aqueous solution of `HCI(pH=2.0)` is mixed with 300 mL of an aqueous solution of `NaOH(pH=12.0)` ?

Explanation

pH of HCI solution `=2,[H_(3)O^(+)]` in solution `=10^(-2)M`
Number of moles of `[H_(3)O^(+)]` in 200 mL of HCI solution `=(10^(-2) xx 200)/(1000) = 0.2 xx 10^(-2) mol`
pH of NaOH solution =12 , `[OH^(-)]` in solution `= 10^(-2) M`
Total volume of the solution after mixing `=(200 +300) = 500 mL `
Number of moles of NaOH left in the solution after neutralisation
`=(0.3 xx 10^(-2) -0.2 xx 10^(-2)) =0.1 xx 10^(-2) mol.`
`:. [OH^(-)]` in the resulting solution `=0.1 xx 10^(-2) xx (1000)/(500) =2.0 xx 10^(-3) M`
`[H_(3)O^(+)] =(K_(w))/([OH^(-)]] =((10^(-14) M^(2)))/((2.0 xx 10^(-3)M)) =5 xx 10^(-12) m`
`pH =- log [H_(3)O^(+)] =- log (5.0 xx 10^(-12))`
`=(12 - log 5) = (12 - 0.69897) = 11.30103`