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)` ?

Answer

A, C

Explanation

For `HCI`,
`pH = 2, [H^(o+)] = 10^(-2)`
mEq of `HCI = 200 xx 10^(-2) =2`
For `NaOH`,
`pH = 12, pOh = 2, [overset(Theta)OH] = 10^(-2)`
mEq of `NaOH = 300 xx 10^(-2) = 3`
After neutralisation,
mEq of NaOh left `=3 - 2 = 1`
`[overset(Theta)OH] = (1)/(500) =2 xx 10^(-3)`
`pOH = - log [overset(Theta)OH]`
`=3 - log2 = 3 - 0.3010 = 2.699`
`pH = 14 - pOH = 14 - 2699. 11.301`