Question

calculate the % relative error in solubility by using concentration instead of activities for la(io3)3 in 0.05 m kno3

Answer 1

Calculation of solubility using concentration

At equilibrium [La3+]= S and [IO3-]= 3S

Where S=molar solubility

Ksp of la(io3)3 = 7.50 * 10^-11 = [La3+][IO3-]^3 = (S)(3S)^3 = 27S^4

S= molar solubility = 1.3 *10^-3 M

In 0.05 m KNO3,

No common ion is there to affect the solubility of ions

Calculation of solubility using activities

Activity=a=concentration*activity coefficient(y) from tables

Y(La3+)=0.245 (for 0.05 M ionic strength of the solution)

Y(IO3^-)=0.82 (for 0.05 M ionic strength of the solution)

0.05 m or M in aqueous solvent as the strength of K+ and NO3-

Ksp of la(io3)3 = 7.50 * 10^-11 =a(La3+) a(IO3-)^3= [La3+] * Y(La3+) [IO3-]^3 * Y(IO3^-)=[La3+][IO3-]^3 * Y(La3+) * Y(IO3^-)^3=[ La3+][IO3-]^3 * 0.245*(0.82)^3= La3+][IO3-]^3*0.135

Or, Ksp of la(io3)3 = 7.50 * 10^-11= [La3+][IO3-]^3*0.135

Or, 7.50 * 10^-11= [La3+][IO3-]^3*0.135

Or,55.55*10^-11=27S^4

Or,2.06*10^-11=S^4

Or,0.00206*10^-8=S^4

S=0.21*10^-2=2.1*10^-3

%Relative error=estimated value-true value/true value *100=1.3*10^-3-2.1*10^-3/2.1*10^-3*100=0.8*10^-3/2.1*10^-3*100=38.09%