Question

what is the molarity of each ion present in aqueous solutions of the following compounds prepared by dissolivng 28.0 g of each compound in water to make 785 mL of solution?
a. Potassium oxide
b.Sodium hydrogen carbonate
c. Magnesium phosphate

Answer 1

Molarity is nu4mber of moles compound in number volume (C)

M=moles/volume (Concentrations)

K₂О reaction with water: K2O+H2O=2KOH

dissotiation in solution: KOH<=>K⁺+ OH^-

n(K₂O)=1/2n(KOH)=1/2n(K⁺)=/2n(OH^-)

n=28.0/(39*2+16)=0.2978 moles

n(K+)=0.5957

n(OH-)=0.5957

Them molarity is 0.7588

2)NaHCO₃ => Na⁺+HCO3-

n(NaHCO3)=n(Na+)=n(HCO3-)=0.3337

Molarity=0.4251 moles in liter

3)Mg₃(PO₄)₂=3Mg²⁺ +2PO₄^3-

n(Mg3PO4)2)=>1/3*n(Mg2+)=1/2*n(PO43-)

n=0.1065

n(Mg)=0.3195

n(PO4)=0.213

Molarity=0.407

Molarity=0.2713