1.

For the equilibrium : HCO_(3)^(-)hArr H^(+)+CO_(3)^(2-)K=4.8xx10^(-11), [CO_(3)^(2-)]=1.1xx10^(-3)M, [HCO_(3)^(-)]=9.8xx10^(-3)MThe pH of the solution is :

Answer»

`8.37`
`9.37`
`6.0`
`8.0`

SOLUTION :`K=([H^(+)][CO_(3)^(2-)])/([HCO_(3)^(-)])`
or `[H^(+)]=(K[HCO_(3)^(-)])/([CO_(3)^(2-)])`
`=(4.8xx10^(-11)xx9.8xx10^(-2))/(1.1xx10^(-3))`
`=4.28xx10^(-9)`
`pH=-LOG[H^(+)]=-log (4.28xx10^(-9))`
`=8.37`


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