A 500 mL saturated solution of MgCO 3 (M M = 84.3 g/mol) is reduced to 120 mL by

Question

A 500 mL saturated solution of MgCO3 (MM = 84.3 g/mol) is reduced to 120 mL by evaporation. What mass of solid MgCO3 is formed? Ksp = 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 Ksp=s2 and s=Ksp1/2 (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 MM to get the mass.

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

Can anyone confirm this?

Explanation / Answer

V = 500 ml

Sat solution MgCO3

MW = 84.5 g/gmol

V2 = 120 ml

Mass of MgCO3 if Ksp = 4*10^-5

MgCO3 <-> Mg+2 and CO3-2

Ksp = [Mg+][CO3-]

assume S is solubility... Since Mg+ : MgCO3 and CO3- : 1 MgCO3

Ksp = S*S

Ksp = S^2

So actualy, the ksp, is the second power of the solbility

S = sqrt(KSp) = sqrt(4.0*10^-5) = 0.006324

Solubility (mol per liter) = 0.006324

Calculate total amount of moles in 500 ml

M= moles / V

0.006324 = moles / 0.5

moles = 0.003162 of MgCO3 in 500 ml

now calculate amount of soluble MgCO3 in 120 ml

M = moles/mole

0.006324 = mol / 0.12

mol = 0.0009486

IF we got previously

0.003162 in solution and then evaporate and 0.0009486 are left

solid (crystal) MgCO3 must be = 0.003162 - 0.0009486 = 0.022145 mol

but we want mass

So

m= moles * MW = 0.022145*84.4 = 1.869 grams of MgCO3 are now solids

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