How much KOH should be dissolved in one litre of solution to prepare a solution having a pH of 12 at `25^@C?`

Explanation

KOH is a strong alkali and is completely dissociated into the constitutent ions,
`KOH+H_(2)O ("excess") to K_(aq)^+ +OH_(aq)^+`
In a solution having pH =12, the hydrogen ion concentration is given by the equation.
pH `=-log [H^+]`
`12=-log[H^+]`
or `[H^+]=10^(-12)"mol L"^(-1)`
Since, the ionic product in water should have a fixed value hence at `25^@C`
`k_(w)=1.0 xx 10^(-14)`
So, `1.0 xx 10^(-14) =[H^+] [OH^(-)]`
Thus, gives, `[OH^(-)]=(1.0 xx 10^(-14))/(10^(-12))=1.0 xx 10^(-2)" mol L"^(-1)`
Since KOH is completely dissociated, hence
`[KOH]=[OH^(-)]=1.0 xx 10^(-2)" mol L"^(-1)`
Molar mass of KOH `=(39+16+1)" g mol"^(-1)="56 g mol"^(-1)`
Then, conc. of KOH `=1.0 xx 10^(-2)" mol L"^(-1) xx "56 g mol"^(-1)`
`=0.56 g L^(-1)`
Thus, 0.56g of KOH should be dissolved per litre solution to obtain a solution of pH 12.