To calculate the mass of solid MgSO4·7H2O needed to precipitate all the PO3- ions, we need to use stoichiometry. The balanced equation for the reaction between MgSO4 and NH4H2PO4 is:
MgSO4 + NH4H2PO4 → MgNH4PO4 + H2SO4
From the equation, we can see that one mole of MgSO4 reacts with one mole of NH4H2PO4 to produce one mole of MgNH4PO4. Therefore, we need to find the number of moles of PO3- ions present in the 10.0 mL of 2.0 M solution of P(PO4)3-.
First, let's find the number of moles of P(PO4)3- ions:
Molarity (M) = moles / volume (L)
So, moles = Molarity × volume (L)
moles of P(PO4)3- ions = 2.0 M × (10.0 mL / 1000 mL/L) = 0.020 moles
Since the stoichiometry of the reaction is 1:1 between MgSO4 and NH4H2PO4, we need the same number of moles of MgSO4 to precipitate all the PO3- ions.
Now, let's calculate the molar mass of MgSO4·7H2O:
MgSO4: 24.305 + 32.065 + (4 × 15.999) = 120.366 g/mol
7H2O: 7 × (2 × 1.008 + 15.999) = 126.126 g/mol
Total molar mass = 120.366 g/mol + 126.126 g/mol = 246.492 g/mol
Now, let's calculate the mass of MgSO4·7H2O needed:
Mass = moles × molar mass
Mass = 0.020 moles × 246.492 g/mol ≈ 4.93 grams
So, approximately 4.93 grams of solid MgSO4·7H2O must be added to precipitate out all of the PO3- ions.
