Given solutions of #(i)# #Al_2(SO_4)_3#; #(ii)# #BaCl_2#; #(iii)# #Na_2SO_4(aq)#; #(iv)# #"equal volumes of (ii) and (iii)"#........which preparation should exert the MOST osmotic pressure?

1 Answer
anor277
Jun 2, 2017

We gots #(i)# #Al_2(SO_4)_3#; #(ii)# #BaCl_2#; #(iii)# #Na_2SO_4(aq)#; #(iv)# #"equal volumes of (ii) and (iii)"#........I suspect it is solution #(iv)#

Explanation:

And #(iv)# a solution obtained by mixing equal volumes of #(ii)# and #(iii)#.

Now the highest osmotic pressure will be expressed by the solution which CONTAINS LEAST number of ions..........

And #(i)# has got 5 ions in solution, #2xxAl^(3+)#, and #3xxSO_4^(2-)#.................

And #(ii)# has got 3 ions in solution, #2xxCl^(-)#, and #Ba^(2+)#.................

And #(iii)# has got 3 ions in solution, #2xxNa^(+)#, and #SO_4^(2-)#.................

But #(iv)# mixes #(ii)# and #(iii)#, and you know, or should know that #BaSO_4# is pretty insoluble stuff. (All sulfates are soluble except for #HgSO_4#, and #CaSO_4#, and #BaSO_4#.

And thus, the following reaction occurs........

#Ba^(2+) + SO_4^(2-)rarr BaSO_4(s)darr#

In this solution, there remain 2 equiv each of #Na^+# and #Cl^-#. As far as I can tell (and you have not set out the problem as clearly as I would like), because #BaSO_4# precipitates, this solution, #(iv)#, has the LEAST number of ions in solution, and thus the highest osmotic pressure.

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