Question

Classify the following ionic compounds according to the Sigma-Aldrich solubility scale based on their molar solubility (S).

Answer choices are: Very Soluble, Freely Soluble, Soluble, Sparingly Soluble, Slightly Soluble, Very Slightly Soluble, and Practically Insoluble.

a) Silver nitrite (AgNO2), S = 0.27 M. Answer is

b) Lithium Phosphate (Li3PO4), S = 3.4 x 10-3 M. Answer is

c) Aluminum Fluoride (AlF3), S = 8.6 x 10-2 M. Answer is

d) Tin (II) Iodide (SnI2), S = 2.9 x 10-2 M. Answer is

Answer 1

The Sigma -Aldrich solubility scale can be given as follows:

Very Soluble (less than 1mL solvent for 1g solute),

Freely Soluble (1 mL to 10 mL solvent for 1g solute),

Soluble (10 mL to 30 mL solvent for 1g solute),

Sparingly Soluble (30 mL to 100 mL solvent for 1g solute),

Slightly Soluble(100 mL to 1000 mL solvent for 1g solute),

Very Slightly Soluble (1000 mL to 10,000 mL solvent for 1g solute),

Practically Insoluble ( more than 10,000 mL solvent for 1g solute)

Thus, given ionic compounds can be classified as shown below:

a) Silver nitrite (AgNO2), S = 0.27 M.

Molar mass of silver nitrite is 154 g/mol.

Thus, 0.27M of AgNO2 = (0.27 mol/1000mL) x (154 g/mol) = 0.04158 g/mL

Therefore, for 1g of AgNO2 solvent used = 1 / (0.04158 g/mL) = 24.05 mL / g

Hence, for 1g of AgNO2 solvent used is around 24mL. Therefore, according to Sigma-Aldrich scale AgNO2 is soluble.

------------------------------------------------------------------------------------------------------------------------------------

b) Lithium Phosphate (Li3PO4), S = 3.4 x 10-3 M.

Molar mass of Li3PO4 is 115.80 g/mol.

Thus, 3.4 x 10-3 M of Li3PO4 = (3.4 x 10-3 mol/1000mL) x (115.80 g/mol) = 0.000394 g/mL

Therefore, for 1g of Li3PO4 solvent used = 1 / (0.000394 g/mL) = 2540 mL / g

Hence, for 1g of Li3PO4 solvent used is around 2540 mL. Therefore, according to Sigma-Aldrich scale Li3PO4 is very slightly soluble.

------------------------------------------------------------------------------------------------------------------------------------

c) Aluminum Fluoride (AlF3), S = 8.6 x 10-2 M.

Molar mass of AlF3 is 84.0 g/mol.

Thus, 8.6 x 10-2 M of AlF3= (8.6 x 10-2 mol/1000mL) x (84.0 g/mol) = 0.007224 g/mL

Therefore, for 1g of AlF3solvent used = 1 / (0.007224 g/mL) = 138.4 mL / g

Hence, for 1g of AlF3 solvent used is around 138.4 mL. Therefore, according to Sigma-Aldrich scale AlF3 is slightly soluble.

------------------------------------------------------------------------------------------------------------------------------------

d) Tin (II) Iodide (SnI2), S = 2.9 x 10-2 M.

Molar mass of SnI2 is 372.5 g/mol.

Thus, 2.9 x 10-2 M of SnI2= (2.9 x 10-2 mol/1000mL) x (372.5 g/mol) = 0.01080 g/mL

Therefore, for 1g of SnI2 solvent used = 1 / (0.01080 g/mL) = 92.6 mL / g

Hence, for 1g of SnI2 solvent used is around 92.6 mL. Therefore, according to Sigma-Aldrich scale SnI2 is sparingly soluble.