pH of a saturated solution of Ba(OH)2 is 12 . The value of solubility product Ksp of Ba(OH)2 is

Dear student

 

pH + pOH = 14pOH = 14 -12 =2 (given pH =12)pOH = -log[OH-][OH-] = 10-pOH= 10-2.............1Ba(OH)2---> Ba+2+ 2OH-At equilibrium:Ba+2= x andOH-= 2xsince 2x =10-2as eq. 1therefore x = 10-2/2 = 0.5 * 10-2ksp= [Ba+2]*[OH-]2= [0.5* 10-2][10-2]2= 0.5 *10-6=5 *10-7