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If the concentration of the weak monoprotic acid HA is C mmol `L^(-1)` and its ionization constant is `K_(a)`, thenA. `C_(H)^(+)=C//2`B. `C_(H)^(+)=sqrt(C )`C. `C_(H)^(+)=sqrt(K_(a)C)`D. `C_(H)^(+)=C//C_(a)` |
Answer» Correct Answer - C `{:(,HA(aq.),hArrH^(+)(aq.),+A^(-)(aq.)),("Inital (M)",C,0,0),("Change (M)",C-Calpha,Calpha,Calpha),("Equilibrium (M)",C-Calpha,Calpha,Calpha):}` where `alpha` is the degree of ionization. `K_(a)=(C_(H^(+))C_(A^(-)))/(C_(HA))` `=((Calpha)(Calpha))/(C(1-alpha))` `=Calpha^(2)` because `alpha lt lt 1` or `alpha=sqrt((K_(a))/(C ))` `C_(H^(+))=Calpha=Csqrt((K_(a))/(C ))=sqrt(K_(a)C )` |
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