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A saturated solution of H_2S in water has concetration of approximately 0.10 M. What is the pH of this soluton and equilibrium concentrations of H_2S,HS^(-) andS^(2-)? Hydrogen sulphide is a diprotic acid and its dissociation constants are K_(a_1)=9.1xx10^(-8) ,K_(a_2)=1.3xx10^(-3) "mol L"^(-1) respectively.

Answer» <html><body><p><br/></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> dissociation <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> is very very small than first dissociationconstant and therefore , the concentration of `H_3O^+` is obtained only from the first dissociation : <br/> `{:(,H_2S+H_2O hArr, H_3O^(+)+, <a href="https://interviewquestions.tuteehub.com/tag/hs-492079" style="font-weight:bold;" target="_blank" title="Click to know more about HS">HS</a>^-),("Initial conc.", 0.10,0,0),("At equilibrium" , 0.10-x, x ,x):}` <br/>(x is amount of `H_2S` dissociated ) <br/>`K_(a_1)=([H_3O^+][HS^-])/([H_2S])=9.1xx10^(-8)` <br/>or `(x xx x)/(0.10-x)=9.1xx10^(-8)` <br/> Solving `x=9.5x10^(-5)` <br/>`therefore [H_3O^+]=[HS^-]=9.5xx10^(-5) "mol L"^(-1)` <br/>`pH=-log [H_3O^+]=-log (9.5xx10^(-5))` <br/> `=-log 9.5 +5 =-0.98+5=4.02` <br/>`[H_2S]=0.10-9.5xx10^(-5)`=0.10 M <br/> To calculate the concentration of `S^(2-)` ion , we are to consider the second dissociation : <br/> `{:(,HS^(-)+H_2O hArr, H_3O^(+)+,S^(2-)),("Initial conc.", 9.5xx10^(-5), 9.5xx10^(-5),0),("At equi.",9.5xx10^(-5)-x, 9.5xx10^(-5)+x ,x):}`<br/>`K_(a_2)=([H_3O^+][S^(2-)])/([HS^-])` <br/> `=((9.5xx10^(-5)+x)xx x)/(9.5xx10^(-5)-x)` <br/> Assuming x to be very very small : <br/> `9.5xx10^(-5)-x <a href="https://interviewquestions.tuteehub.com/tag/approx-882876" style="font-weight:bold;" target="_blank" title="Click to know more about APPROX">APPROX</a> 9.5 xx10^(-5)` and `9.5xx10^(-5)+x` <br/>=`9.5xx10^(-5)` <br/>Solving `x=1.3xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)` <br/>`therefore [S^(2-)]=1.3xx10^(-13)` <br/> Thus, `[H_3O^+]=9.5xx10^(-5)` M , `[HS^-]=9.5xx10^(-5)` M, `[S^(2-)]=1.3xx10^(-13)M , [H_2S]`=0.01 M , pH=4.02</body></html>


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