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Equilibrium

Question
CBSEENCH11006623

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6·3 × 10–18).

Solution
Let the concentration of both FeSO4 and Na2S solutions before mixing = x mol L= xM
The solubility equilibrium of iron sulphide may be represented as,
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Now, Fe2+ ions are to be provided by FeSOand S2– ions and are to be provided by Na2S, as a result of dissociation.
FeSO subscript 4 space rightwards arrow with aq on top space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis
Na subscript 2 straight S space rightwards arrow with aq on top space 2 Na to the power of plus left parenthesis aq right parenthesis space plus space straight S to the power of negative 2 end exponent left parenthesis aq right parenthesis
Since equal volumes of the two solutions have been mixed, the concentration of the solutions as well as of ions on mixing will be reduced to half i.e.
 straight x over 2 straight M
therefore space space space space Before space mixing comma space open square brackets Fe to the power of 2 plus end exponent close square brackets space equals space xM
space space space space space space After space mixing comma space open square brackets Fe to the power of 2 plus end exponent close square brackets space equals space straight x over 2 straight M
Similarly comma space Before space mixing comma space open square brackets straight S to the power of 2 minus end exponent close square brackets space equals space xM
space space space space space space space space After space mixing comma space space open square brackets straight S to the power of 2 minus end exponent close square brackets space equals space straight x over 2 straight M
Ionic space product space space equals space open square brackets Fe to the power of 2 plus end exponent close square brackets space open square brackets straight S to the power of 2 minus end exponent close square brackets space equals straight x over 2 cross times straight x over 2 space equals straight x squared over 4
Since oh mixing, there is no precipitation of iron sulphide, therefore,
Ionic product = Solubility product
straight x squared over 4 space equals space 6.3 space cross times space 10 to the power of negative 18 end exponent
straight x squared space equals space 4 space cross times space 6.3 space cross times space 10 to the power of negative 18 end exponent space equals space 2.52 space cross times space 10 to the power of negative 18 end exponent
straight x space equals left parenthesis 2.52 space cross times space 10 to the power of negative 18 end exponent right parenthesis to the power of 1 divided by 2 end exponent space equals space 5.02 space cross times space 10 to the power of negative 9 end exponent straight M
The maximum concentration of both the solutions is 5·02 × 10–9M