Question

For calcite dissolution experiment, when you increase ionic strength from 0 (infinite diluted) to 0.1 m, how much does the solubility increase？

Answer 1

For calcite dissolution equilibrium exists in solution as follows,

CaCO3--------->Ca2+ +CO3-

ksp(CaCO3)=[Ca2+][CO₃^2-]=S * ^2   ,S=[Ca2+]=[CO32-]=solubility or concentration at equilibrium

When the equilibrium is pushed forward the concentration of Ca2+ increases and hence increases ksp or solubility product or solubility.w1hen you increase ionic strength from 0 (infinite diluted) to 0.1 m,

I=1/2 cizi^2 ,where ci=concentration of ions zi=charge on ions

ksp depends on which can be calculated using Debye-Huckel equation ,hence

log=-[0.509 (Z+ Z-)(I)^1/2]/[1+(I)^1/2]

when I=0 ,low concentration at infinite dilution, =1=unity

So S=ksp=3.3*10^-9 mol/L

At I=0.1m=0.1 mol/kg=0.1mol/L(in aqueous solvent density=1kg/L)

log=-[0.509 (2*2)(0.1)^1/2]/[1+(0.1)^1/2]=-[2.35*]/[1.32]=-1.78

=0.016

So,

ksp(CaCO3)=S * ^2  

ksp(CaCO3)=3.3*10^-9

S=3.3*10^-9 /0.016=206.25*10^-9 mol/L

change in solubility=206.25*10^-9 mol/L-3.3*10^-9 mol/L=202.95*10^-9M (increase by this amount)