Estimate the solubility of TISCN in a solution of which ionic strength is 1.265e-2M at 25 degree and Ksp=1.6e-4.
I do not know why Prof. gives me ionic strength? In the next question, Prof. said " For a sparingly soluble ionic compound of type AB the solubility can be calculated using the following relationship, solubility S= (Ksp) ^(1/2). Si S=(Ksp TISCN) ^(1/2) Give a qualitative explanation for the difference in solubility calculated by both ways. Any help will be greatly appreciated.
5 answers
what you must do is calculate the solubility of TiSCN as if concn in M is the same as activity. In the other case you want to use ionic strength to calculate activity of Ti and activity of SCN and use that to calculate solubility. Compare the solubility by the two methods.
Thank you DrBob222. My activity of Ti is 0.8661 and SCN is 0.8896. I do not know any equation between activity and solubility. Can you help me with that? Thank you
Note: the first post noted Tl, not Ti.
Ksp = (Tl^+)(SCN^-) and the activity is supposed to be substituted directly; i.e.
Ksp = aTl^+ x aSCN^-. Since you know activity of each(if those numbers are really activities), substitute those and solve for solubility just as you did for the first one.
However, I wonder how you obtained the activity of each without any other date. Since the activity is concn x activity coefficient, I wonder if that 0.8661 and 0.8896 are activity coefficients and not activity. I think they must be activity coefficients. If they are then you do this instead of the above.
Ksp = (Tl^+)(SCN^-)
Ksp = (Tl^+)*0.8661*(SCN^-)*0.8896
Ksp = 0.8661x*0.8896x
and solve for x. If you have further questions please clarify those numbers as to whether they are activities or activity coefficients.
Ksp = (Tl^+)(SCN^-) and the activity is supposed to be substituted directly; i.e.
Ksp = aTl^+ x aSCN^-. Since you know activity of each(if those numbers are really activities), substitute those and solve for solubility just as you did for the first one.
However, I wonder how you obtained the activity of each without any other date. Since the activity is concn x activity coefficient, I wonder if that 0.8661 and 0.8896 are activity coefficients and not activity. I think they must be activity coefficients. If they are then you do this instead of the above.
Ksp = (Tl^+)(SCN^-)
Ksp = (Tl^+)*0.8661*(SCN^-)*0.8896
Ksp = 0.8661x*0.8896x
and solve for x. If you have further questions please clarify those numbers as to whether they are activities or activity coefficients.
Thank you DrBob222. Yes, 0.8661 and 0.8896 are activity coefficients. That is my bad. Thanks to your suggestion, I already figure out the solubility of TISCN is 0.01425. Besides, the solubility would be 0.01265 if I applied the formula S=Ksp ^(1/2). What makes the big difference between these two numbers? The error between them are kind of big
That's exactly right and I agree with both of the values you have. The profs purpose in assigning a problem like this is to illustrate just how much the solubility increases when one takes into account the activity coefficients. You see when you use molarity by itself you are assuming the activity coefficient is 1.00 so that activity = molarity x 1.00. But when the activity coefficient is not 1.00 (and it is hardly ever 1.00 -- ONLY in VERY dilute solutions is it 1.00) then activity = M x fraction less than 1 which makes the activity less than the molarity. You can see from my bold face work above that the effect of the activity coefficent is to increase Ksp.
Note (Ksp/0.8661*0.8896) = x^2 and that increases x. The error is significant in some cases and not so bad in others. The error is ALWAYS worse in more concentrated solutions because the activity coefficient is smaller. And when you divide Ksp by smaller and smaller numbers the result is larger and larger numbers and that increases x = solubility. So why do we almost never use activities and activity coefficients. Mostly because they are very inconvenient and time consuming AND because determining activity coefficients is not all that easy. In addition they change under normal circumstances and that makes it even more inconvenient. So I think we use the simple method and realize it's a close approximation. The difference usually is not this pronounced but the prof chose an example where Ksp is large already and it makes more difference. If you had something like CuS with Ksp way down there in the 10^-45 or so, I don't think you would see very much difference. Glad to be of help.
Note (Ksp/0.8661*0.8896) = x^2 and that increases x. The error is significant in some cases and not so bad in others. The error is ALWAYS worse in more concentrated solutions because the activity coefficient is smaller. And when you divide Ksp by smaller and smaller numbers the result is larger and larger numbers and that increases x = solubility. So why do we almost never use activities and activity coefficients. Mostly because they are very inconvenient and time consuming AND because determining activity coefficients is not all that easy. In addition they change under normal circumstances and that makes it even more inconvenient. So I think we use the simple method and realize it's a close approximation. The difference usually is not this pronounced but the prof chose an example where Ksp is large already and it makes more difference. If you had something like CuS with Ksp way down there in the 10^-45 or so, I don't think you would see very much difference. Glad to be of help.
1 answer