the Kap for Ag2SO4 is 1.5*10^-5. assume unit activity coefficients and determine the molar solubility?

Question

of Ag2SO4
A) in water
B)in 0.50 M Na2SO4
C)in 1.0 mM Na2SO4
Repeat these solubility determinations

D)....using the DHLL to determine the activity coefficients in water
E) ...using the Davis equation to determine the activity coefficients in 0.50M Na2SO4

F) ...using the EDHE to determine the activity coefficients in 1.0mM Na2SO4

Answer 1

I saw your questions and they are pretty advanced and demanding... Where did you get these problems from?

Ag2SO4 <=> 2Ag(+) + SO4(-2)
s .. .. .. .. .. .. .. .. 2s .. .. .. ..s

for γ=1 Kap=Ksp

A)Ksp = [Ag+]^2[SO4(-2)] =(2s)^2*s= 4s^3= 1.5*10^-5 =>
s=1.553* 10^-2 =1.6*10^-2 M (very advanced problems need the answer given in the appropriate significant figures and here they give concentrations expressed with 2 sig. figures)

B)Ksp= (2s)^2*(0.5+s)=> 4s^3+2s^2-1.5*10^-5=0 =>
s=2.7* 10^-3 M
using the calculator at http://www.1728.com/cubic.htm

c) Ksp=4s^2 (0.001+s)=> 4s^3 + 0.004s^2 -1.5*10^-5=0
=> s=1.5 *10^-2 M

D) I=1/2 *([Ag+]*1^2+[SO4(-2)]*2^2) = 1/2*(2s+4s)=3s
-log γ(+) = 0.512*Sqrt(3s)=> γ(+)=10^-0.512sqrt(3s)
-log γ(-)= 0.512*2*sqrt(3) => γ(-)=10^-1.024sqrt(3s)

Ka=γ(+)^2γ(-) [Ag+]^2 [SO4(-2)]=
=10^-2*0.512sqrt(3s) *10^-1.024sqrt(3s)*4s^2*s=
=10^[-2.048 *sqrt(3s)]*4s^3 =>

s=2.4*10^-2 M if the calculator at http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic
worked fine

e)I=1/2 (0.5*2*1^2 + 0.5*2^2+ 2s+4s)= 1.5+3s

-log γ(+/-) = 0.512*1*2*[sqrt(I)/(sqrt(I)+1) -0.2I]

Kap= γ(+/-) ^(1+2)*4s^3= γ(+/-)^3*4s^3

Good luck solving this... even if you approximate and say that the ionic strength is determined only by Na2SO4 then it is easier, but you can't do this approxiamation for f...

(f) -log γ(+/-) =0.512*1*2*sqrt(I) / (1+0.33*α*sqrt(I))

You need α...

If your tutor gives you the solutions please send them to me (even from the other questions) I am really curious to see how they solve them/