Question

For Pb(NO3)2, the fraction of Pb2+ ions that associate with NO3 - ions to form ion pairs (PbNO3 + ) is known to be equal to 43% in a 0.100 mol/kg aqueous solution at 25 oC. Calculate the ionic strength of this solution.

Answer 1

[Pb2+]=concentration=0.1 mol/Kg=0.1 mol/1000g of water=0.0001 mol/g

density of water at 25 degrees celsius =0.997g/ml

so 1g water=0.997 ml

[Pb2+]=0.0001 mol/0.997 ml=0.0001 mol/ml =0.0001 mol/10^-3 L=0.1 mol/L

[Pb2+]=0.1 mol/L

[NO3^2-]=43% *0.100 mol/kg=43/100*0.1 =0.043 mol/kg =0.043 mol/1000g

density of water at 25 degrees celsius =0.997g/ml

[NO3^2-]=0.043 mol/0.997g/ml =0.043 *10^-3 mol/ml=0.043 *10^-3 mol/1000L=0.043 mol/L

Now ,I=ionic strength=1/2 ci*zi²   where ci=molar concentration of ion in mol/L and zi=charge on that ion

I=1/2 ([Pb2+] * (# of ions ) *charge +[NO32-] *(# of ions ) *charge]=1/2(0.1*1*2+0.043*2*2)=0.186 mol/L

I=0.186=0.2 mol/L