If the reaction of phosphate ion with water is ignored, what is the total concentration of ions in a solution prepared by dissolving 7.00 g of K3PO4 in enough water to make 350. mL of solution?

0.377 M

Explanation:

Given that:

mass of = 7.00 g

the molar mass of  = 212.27 g/mol

number of moles =

number of moles of =

= 0.0330

The dissociation of K₃PO₄(aq) → 3K⁺(aq) + PO₄³⁻(aq)

So number of Ions formed = 3 K + and 1 PO₄³⁻ = 4 ions

No of moles of ions = 4 × 0.0330

Molarity = 4 × 0.0330/ ( 350/1000)  

= 0.377 M

Hence, the   total concentration of ions in a solution prepared by dissolving  7.00 g of K3PO4 in enough water to make 350. mL of solution = 0.377 M

the answer for sodium is 0.186

the answer for rubidium is 0.251