Question

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?

Answer 1

For `"b."` the answer is `2.5*mol` of solute ions........

Explanation:

`Cr_2(SO_4)_3(s) rarr 2Cr^(3+) + 3SO_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 `"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 `"biphosphate, "HPO_4^(2-)`:

`Rb_3PO_4(s) + "excess water"rarr3Rb^(+) + HPO_4^(2-) + HO^-`

This question thus was not well-proposed.