What will be the resultant pH when 200 mL of an aqueous solution of HCI (pH = 2) is mixed with 300 mL of an aqueosu solution of NaOH (pH = 12)?

Explanation

pH of HCl =2 `therefore [H_3O^+]=10^(-2)` M
Moles of `H^+` ions in 200 mL of `10^(-2)` M HCl =`10^(-2)/1000xx200=2xx10^(-3)`
pH of NaOH =12
`therefore [H^+]=10^(-12)` or `[OH^-]=10^(-2)`
Moles of `OH^-` ions in 300 ml of `10^(-2)` M NaOH
`=10^(-2)/1000xx300=3xx10^(-3)`
`2xx10^(-3)` mol of `H^+` will be neutralised with `2xx10^(-3)` mol of `OH^-` ions.
Moles of `OH^-` left in 500 ml of solution after mixing
`=3xx10^(-3)-2xx10^(-3)`
`=1xx10^(-3)`
Molar concentration of `OH^-` in resulting solution
`=(1xx10^(-3))/500 xx1000`
`=2xx10^(-3)` M
pOH=-log `(2xx10^(-3))`
=2.698
`therefore` pH=14-2.698 =11.302