# 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?

Jun 2, 2017

We gots $\left(i\right)$ $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$; $\left(i i\right)$ $B a C {l}_{2}$; $\left(i i i\right)$ $N {a}_{2} S {O}_{4} \left(a q\right)$; $\left(i v\right)$ $\text{equal volumes of (ii) and (iii)}$........I suspect it is solution $\left(i v\right)$

#### Explanation:

And $\left(i v\right)$ a solution obtained by mixing equal volumes of $\left(i i\right)$ and $\left(i i i\right)$.

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

And $\left(i\right)$ has got 5 ions in solution, $2 \times A {l}^{3 +}$, and $3 \times S {O}_{4}^{2 -}$.................

And $\left(i i\right)$ has got 3 ions in solution, $2 \times C {l}^{-}$, and $B {a}^{2 +}$.................

And $\left(i i i\right)$ has got 3 ions in solution, $2 \times N {a}^{+}$, and $S {O}_{4}^{2 -}$.................

But $\left(i v\right)$ mixes $\left(i i\right)$ and $\left(i i i\right)$, and you know, or should know that $B a S {O}_{4}$ is pretty insoluble stuff. (All sulfates are soluble except for $H g S {O}_{4}$, and $C a S {O}_{4}$, and $B a S {O}_{4}$.

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

$B {a}^{2 +} + S {O}_{4}^{2 -} \rightarrow B a S {O}_{4} \left(s\right) \downarrow$

In this solution, there remain 2 equiv each of $N {a}^{+}$ and $C {l}^{-}$. As far as I can tell (and you have not set out the problem as clearly as I would like), because $B a S {O}_{4}$ precipitates, this solution, $\left(i v\right)$, has the LEAST number of ions in solution, and thus the highest osmotic pressure.