Calculate the pH of `1.5xx10^(-3)` M solution of `Ba(OH)_2`.

Explanation

`{:(Ba(OH)_(2)" "rarr" "Ba^(2+)+2OH^(-)),(1.5xx10^(-3)M" "2xx1.5xx10^(-3)M):}`
`[OH^(-)]=3xx10^(-3)M`
`[because pH+pOH=14]`
`pH=14-pOH`
`pH=14-(-log[OH^(-)])`
`=14+log[OH^(-)]`
`=14+log(3xx10^(-3))`
`=14+log3+log10^(-3)`
`=14+0.4771-3`
`=11+0.4771`
`pH=11.48`