# What molar quantity of ions are realized in solution when a 0.50*mol quantity of Cr_2(SO_4)_3 is dissolved in water? What about when a 0.50*mol quantity of Rb_3(PO_4)_3 is dissolved?

Feb 4, 2017

For $\text{b.}$ the answer is $2.5 \cdot m o l$ of solute ions........

#### Explanation:

$C {r}_{2} {\left(S {O}_{4}\right)}_{3} \left(s\right) \rightarrow 2 C {r}^{3 +} + 3 S {O}_{4}^{2 -}$

Thus 5 moles of ions result from one mole of solute. We started with half a mole of solute, and thus 2.5 mole ions result.

For $\text{rubidium phosphate}$ we could proceed in the same way. However, there is a little difficulty. Dissolve a phosphate in aqueous solution, and the majority species in solution is $\text{biphosphate, } H P {O}_{4}^{2 -}$:

$R {b}_{3} P {O}_{4} \left(s\right) + \text{excess water} \rightarrow 3 R {b}^{+} + H P {O}_{4}^{2 -} + H {O}^{-}$

This question thus was not well-proposed.