Question

A 500 mL saturated solution of MgCO₃ (M_M = 84.3 g/mol) is reduced to 120 mL by evaporation. What mass of solid MgCO₃ is formed? K_sp = 4.0 x 10^-5 for MgCO₃.

EDIT: If ions can evaporate in water then the answer below is right. If not, this is how I solved the problem.

MgCO₃ dissolves at a 1:1 ratio, therefore K_sp=s² and s=K_sp^1/2 (or sqrt of Ksp)
s = (4.0x10^-5)^1/2 = .00632 M

Using the equation M₁V₁=M₂V₂

M₁ = s = 0.00632; V₁ = 0.500; V₂ = 0.120

Solving for M₂ = 0.0263 M
Then multiply M₂ by M_M to get the mass.

Therefore, mass = (0.0263)(84.3) = 2.22g

Can anyone confirm this?

Answer 1

The part about solubility of MgCO₃ (s = 0.00632 M) is absolutely correct.

This solubility remains constant and does not depend upon volume of solution.

So, whether the solution has volume 500 mL or 120 mL, solubility will not change.

For 500 mL

moles of MgCO₃ dissolved in solution = (solubility of MgCO₃) * (volume of solution in Liter)

moles of MgCO₃ dissolved in solution = (0.00632 M) * (500 x 10^-3 L)

moles of MgCO₃ dissolved in solution = 0.00316 mol

For 120 mL

moles of MgCO₃ dissolved in solution = (solubility of MgCO₃) * (volume of solution in Liter)

moles of MgCO₃ dissolved in solution = (0.00632 M) * (120 x 10^-3 L)

moles of MgCO₃ dissolved in solution = 0.000759 mol

moles of MgCO₃ precipitate out = (initial moles of MgCO₃) - (moles of MgCO₃ in 120 mL)

moles of MgCO₃ precipitate out = (0.00316 mol) - (0.000759 mol)

moles of MgCO₃ precipitate out = 0.002401 mol

mass of MgCO₃ formed = (moles of MgCO₃ precipitate out) * (molar mass MgCO₃)

mass of MgCO₃ formed = (0.002401 mol) * (84.3 g/mol)

mass of MgCO₃ formed = 0.202 g