Question

What is the concentration of Mg²⁺ if 5528.4g of Mg₃P₂ are added to 125L of water?

Ksp(ZnCO3)=1.5*10^-6

Answer 1

first find the moles of Mg3P2

moles = mass / molar mass

= 5528.4 g / 134.8625 g/mol

= 41.0 mol

lets write th edissociated equation of Mg3P2

Mg₃P₂ ---> 3Mg²⁺ + 2P^3-

from this equation it is clear that one mole of Mg3P2 is having 3 mol of Mg2+

accordingly

41.0 mol fo Mg3P2 will have 3 x 41.0 mol   = 123.0 mol

now

we know the moles of Mg2+ and total volume

use formula

Molarity (M) = moles of Mg2+ / volume in liters

M = 123.0 mol / 125 L

M = 0.984 mol /L