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For the equilibrium system : 2HCl(g)...

For the equilibrium system :
`2HCl(g)hArr H_(2)(g)+Cl_(2)(g)`
the equilibrium constant is `1.0xx10^(-5)`. What is the concentration of HCl if the equilibrium concentration of `H_(2)` and `Cl_(2)` are `1.2xx10^(-3)` M and `1.2xx10^(-4)` M respectively ?

A

`12xx10^(-4)M`

B

`12xx10^(-3)M`

C

`12xx10^(-2)M`

D

`12xx10^(-1)M`

Text Solution

Verified by Experts

The correct Answer is:
C
`K=([H_(2)][Cl_(2)])/([HCl]^(2))`
`1.0xx10^(-5)=((1.2xx10^(-3))(1.2xx10^(-4)))/([HCl]^(2))`
or `[HCl]^(2)=((1.2xx10^(-3))(1.2xx10^(-4)))/(1.0xx10^(-5))`
`=1.44xx10^(-2)`
`[HCl]=sqrt(1.44xx10^(-2))=12xx10^(-2)`
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