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For the equilibrium : HCO(3)^(-)hArr ...

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)M`
The pH of the solution is :

A

`8.37`

B

`9.37`

C

`6.0`

D

`8.0`

Text Solution

Verified by Experts

The correct Answer is:
A
`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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