Calculate the pH of the followings :
`0.001 M Ba(OH)_(2)`

Explanation

`Ba(OH)_(2)` ionises as follows :
`Ba(OH)_(2)(aq) hArr Ba^(2+)(aq) + 2OH^(-)(aq)`
Since one mole of `Ba(OH)_(2)` gives 2 moles of `OH^(-)` ions.
`therefore [OH^(-)] = 2[Ba(OH)_(2)] = 2 xx 0.001 = 0.002 M`
`[H_(3)O^(+)] = (K_(w))/([OH^(-)]) = (1 xx 10^(14))/(0.002) = 5 xx 10^(-12)`
`pH = -log [H_(3)O^(+)] = -log [5 xx 10^(-12)] " " [because log 5 = 0.699]`
`= -log 5 + 12 log 10 = -0.699 + 12 = 11.301`