The pH of 0.001 M `Ba(OH)_(2)` solution will be

Explanation

` Ba( OH)_2 ` ionises as
`Ba(OH)_2 (aq) to Ba^(2+) (aq) + 2OH^(-)(aq)`
one mole of Ba `(OH)_2` gives two moles of `OH^(-)` ion। If `Ba(OH)_2` is assumed to be completely ionised we, have
`" " [OH^(-)] =2 xx [Ba(OH)_2 ]= 2 xx 0.001`
` " " = 2.0 xx 10 ^(-3) " mol " L^(-1)`
` therefore " " [H_3O^(+) ] =(K_W)/([OH^(-)])`
` " " = (1.0 xx 10 ^(-14))/( 2.0 xx 10 ^(-3))`
` = 5.0 xx 10 ^(-12) " mol " L^(-1)`
Hence ` " " pH = - log _(10) [H_3O^(+) ]`
` " " = -log _(10) (5.0 xx 10 ^(-12)) `
` " " = 11.30 .`