1.

The K_(sp)ofBaCrO_(4) is 2.4xx10^(-10)M^(2). The maximum concentration of Ba(NO_(3))_(2)possible without precipitation in a 6xx10^(-4)M K_(2)CrO_(4) solution is

Answer»

`4xx10^(-7)M`
`1.2xx10^(10)M`
`6XX10^(-4)M`
`3xx10^(-4)M`

Solution :`K_(sp)=[Ba^(2+)][CrO_(4)^(2-)]`
`2.4xx10^(-10)=[Ba^(2+)](6xx10^(-4))`
`therefore [Ba^(2+)]=(2.4xx10^(-10))/(6xx10^(-4))`
`=4.0xx10^(-7)M`.


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