Question

Including activity coefficients, find the [IO₃^-] in saturated Ba(IO₃)₂ in (a) water, (b) 0.050 M NaOH, and (c) 0.050 M Ba(OH)₂.

Answer 1

Ignore Activity Coefficients Ksp = [Ba2+] [IO3 - ]^2 Ba(IO3)2

Ba2+ 2 IO3 - I

lots 0 0 C - S + S + 2S E lots S 2S So, Ksp = (S) (2S)2 = 1.57 x 10-9

Solving,

S = 7.32 x 10-4 M

Include Activity Coefficients First Determine the Ionic Strength:

 = ½ ( cBa zBa 2 + cIO3 zIO3 2 )

Now we have a problem. We do not know these concentrations. So, we have to estimate them by assuming we can ignore the activity coefficients and use the above result for the solubility.

 = ½ (7.32 x 10-4 x (2)2 + 2 x 7.32 x 10-4 x (1)2 ) = 0.00220 M

Now, Determine the Activity Coefficients (using the Debye-Huckel model):

log Ba = - (0.51) (2)2 = - 0.09568 Ba = 10-0.09568 = 0.802 log IO3 = - (0.51) (1)2 = - 0.02392 IO3 = 10-0.02392 = 0.946

So, Ksp = Ba2+ [Ba2+] IO3 2 [IO3 - ] 2 = (0.802) (S) (0.946)2 (2S)2 = 1.57 x 10-9

Solving, S = 8.18 x 10-4 M

Now we have a better approximation for the concentration of the ions, so we can re-calculate the Ionic Strength:

 = ½ (8.18 x 10-4 x (2)2 + 2 x 8.18 x 10-4 x (1)2 ) = 0.00245 M

Re-calculate the Activity Coefficients:

log Ba = - (0.51) (2)2 = - 0.1010 Ba = 10-0.1010 = 0.792 log IO3 = - (0.51) (1)2 = - 0.02524 IO3 = 10-0.02524 = 0.944 So, Ksp = Ba2+ [Ba2+] IO3 2 [IO3 - ] 2 = (0.792) (S) (0.944) 2 (2S)2 = 1.57 x 10-9

Solving, S = 8.22 x 10-4 M

We could continue to iterate, but the solubility did not change much after the 2nd iteration. So, at this point, we will consider the problem solved. For Comparison:

without 's including 's S = 7.32 x 10-4 M 8.22 x 10-4 M Percentage Error in not Including Activity Coefficients: % Error = ( | 8.22 - 7.32 | / 8.22 ) x 100 = 11%

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