pH of an aqueous solution of 25ºC is 12. On increasing its dilution, at 25ºC, upto 10 times its new pH would be :-

(1)

13

(2)

120

(3)

10

(4)

11

Ph=14-Poh and Poh = -log(concentration of OH ion) here the concentration is decreased by a factor of 10 hence new Poh` = Poh +1 =2+1=3 hence new Ph=11 please do approve my answer if correct