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?

1 Answer
Feb 4, 2017

Answer:

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.