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For the following reaction, equilibrium ...

For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)`
`Fe_((aq))^(2+)+S_((aq))^(2) hArr FeS(s)`
When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is

Text Solution

Verified by Experts

The correct Answer is:
8.93
`pH=1/2pK_a + 1/2pK_a+1/2 log C`
`8.2=1/2(14)+1/2(3.4)+1/2log C`
`1/2` log C =8.2-7-1.7=-0.5
log C =- 0.5 x 2 =-1
`therefore` C=0.1
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