1.

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

Answer»

Solution :`{:(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`


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