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.

Related questions
See all questions in Osmolarity
Impact of this question
2142 views around the world
You can reuse this answer
Creative Commons License