Calculate Solubility Product of Inorganic Ionic Compounds and Solubility

Solubility product and solubility are two key factors used to express soluble amount (or quantity) of an inorganic ionic compound in water. As well as, by looking those values we can decide though a compounds is soluble or not soluble in water. In this tutorial, we use solubility product and solubility value in different calculations to decide solubility of compounds. However, our main focus of this tutorial is how to calculate solubility product (Ksp) from solubility value.


Content

  • Definition of solubility
    • Calculate solubility with examples
  • Definition of solubility product (Ksp)
    • Calculate solubility product (Ksp) for ionic inorganic compounds
    • Calculate solubility product when solubility value is given for a specific ionic inorganic compound
      • Sodium chloride (NaCl)
      • Silver chloride (AgCl
      • Lead chloride (PbCl2)
    • Solubility product values for some inorganic ionic compounds
  • Deciding the solubility of a compound by observing solubility value and solubility product
  • Solubility value and solubility product variation against temperature
  • Questions


Definition of solubility

There is a maximum quantity (g) dissolving in a unit volume or quantity of water (or in another solvent) for a specific compound. As an example, maximum quantity of dissolving Sodium chloride (NaCl) is much higher than maximum quantity of dissolving Silver chloride (AgCl) in unit volume of water at same temperature. Unit of solubility can be expressed in g/100 ml of water or g/100 g of water or in other unit as required. These units are important in calculations.

You can see a significant difference in solubility of NaCl and AgCl. Usually solubility increases with temperature.



Calculate solubility with examples



Example 1:

Solubility of Sodium chloride is 36.0 g/100 g of water at 250C. How much sodium chloride can be dissolved in 500g of water.


Answer: This is a very simple question.

  • Solubility of NaCl in 1g of water = 36.0 g / 100 g of water
  • Solubility of NaCl in 1g of water = 0.36 g
  • NaCl Quantity of dissolving in 500 g of water = 0.36 g / 1g of water * 500 g of water
  • NaCl Quantity of dissolving in 500 g of water = 180.0 g


Definition of solubility product (Ksp)

The solubility product constant (Ksp), is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp value it has.

Solubility product concept is mostly used for insoluble inorganic ionic compounds. Solubility product is also a constant at room temperature.


Write the solubility product equation for Silver chloride as below.


Step 1: First write the balanced equation.

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)


Step 2: Write the Ksp equation

Ksp, AgCl = [Ag+(aq)] [Cl-(aq)]

  • [Ag+(aq)] = concentration of Ag+ cation in the aqueous solution
  • [Cl-(aq)] = concentration of Cl- anion in the aqueous solution


Write the solubility product equation for Lead chloride as below.


Step 1: First write the balanced equation.

PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)


Step 2: Write the Ksp equation

Ksp, PbCl2 = [Pb2+(aq)] [Cl-(aq)]2

  • [Pb2+(aq)] = concentration of Pb2+ cation in the aqueous solution
  • [Cl-(aq)] = concentration of Cl- anion in the aqueous solution


Example 2

Find solubility product of NaCl

Solubility of NaCl is 36.09 g / 100 ml of water at 200C. Calculate the solubility product.


Answer

Maximum 36.09 g of NaCl can be dissolved in 100 ml of water. To find the solubility product, concentrations (mol dm-3) of Na+ and Cl- should be found when 36.09 g of NaCl is dissolved in water.


Step 1: First, we are going to find, dissolving mass of NaCl in one liter (1 dm3) of water.

  • Maximum mass of dissolving NaCl in one liter (1dm3) = 36.09 * (1000/100) g
  • Maximum mass of dissolving NaCl in one liter (1dm3) = 360.9 g

Step 2: Now we know the maximum mass of dissolving NaCl in 1 dm3 of water. Next step is, find the dissolved amount of moles in 1 dm3 of water.


Molecular mass of NaCl = 23 + 35.5

Molecular mass of NaCl = 58.5 g mol-1


  • maximum amount of dissolving NaCl in one liter (1dm3) of water = 360.9 g / 58.5 g mol-1
  • maximum amount of dissolving NaCl in one liter (1dm3) of water = 6.17 mol

Maximum concentration of NaCl can be dissolved in one liter (1dm3) of water = 6.17 mol dm-3



Step 3: Now, we are going to calculate solubility product of NaCl as following (solubility product equation for NaCl).

solubility product equation of NaCl

Step 4: Now, we need to find the concentrations of Na+ ion and Cl- ions in water.

NaCl completely dissociate to Na+ ion and Cl- ions in water.

NaCl(aq) → Na+(aq) + Cl-(aq)


[Na+(aq)] = 6.17 mol dm-3

[Cl-(aq)] = 6.17 mol dm-3


Ksp(NaCl) = [Na+(aq)] * [Cl-(aq)]

Ksp(NaCl) = 6.17 mol dm-3 * 6.17 mol dm-3

Ksp(NaCl) = 38.07 mol2 dm-6


We can see, Ksp of NaCl is very high. Therefore it indicates NaCl dissolves in water in higher amounts.



Example 3

Find solubility product of Silver chloride (AgCl)

Solubility of AgCl is 0.00052 g/100 g of water at 500C. Calculate the solubility product.


Answer

Maximum 0.00052 g of AgCl can be dissolved in 100 ml of water. To find the solubility product, concentrations (mol dm-3) of Ag+ and Cl- should be found when 0.000520 g of AgCl is dissolved in water.


Step 1: First, we are going to find, dissolving mass of AgCl in one liter (1 dm3) of water.

  • Maximum mass of dissolving AgCl in one liter (1dm3) = 0.000520 * (1000/100) g
  • Maximum mass of dissolving AgCl in one liter (1dm3) = 0.0052 g

Step 2: Now we know the maximum mass of dissolving 0.0052 g in 1 dm3 of water. Next step is, find the dissolved amount of AgCl moles in 1 dm3 of water.


Molecular mass of AgCl = 107.9 + 35.5

Molecular mass of AgCl = 143.4 g mol-1


  • Maximum amount of dissolving AgCl in one liter (1dm3) of water = 0.0052 g / 143.4 g mol-1
  • Maximum amount of dissolving AgCl in one liter (1dm3) of water = 0.000036 mol

Maximum concentration of AgCl can be dissolved in one liter (1dm3) of water = 0.000036 mol dm-3



Step 3: Now, we are going to calculate solubility product of AgCl as following (solubility product equation for AgCl).

Ksp, AgCl = [Ag+(aq)] [Cl-(aq)]



Step 4: Now, we need to find the concentrations of Ag+ ion and Cl- ions in water.

AgCl poorly dissociate to Ag+ ions and Cl- ions in water.

AgCl(aq) → Ag+(aq) + Cl-(aq)


[Ag+(aq)] = 0.000036 mol dm-3

[Cl-(aq)] = 0.000036 mol dm-3


Ksp(AgCl) = [Ag+(aq)] * [Cl-(aq)]

Ksp(AgCl) = 0.000036 mol dm-3 * 0.000036 mol dm-3

Ksp(AgCl) = 1.296 * 10-9 mol2 dm-6


We can see, Ksp of AgCl is very low compared to Ksp value of NaCl. That gives us an indicate NaCl is much soluble in water than AgCl.



Example 4

Find solubility product of Lead chloride (PbCl2)

Solubility of PbCl2 is 0.99 g/100 mL of water at 200C. Calculate the solubility product and mention assumptions made by you during calculations.


Answer

Maximum 0.99 g of PbCl2 can be dissolved in 100 ml of water. To find the solubility product, concentrations (mol dm-3) of Pb2+ and Cl- should be found when 0.99 g of PbCl2 is dissolved in water.


Assumptions made during the calculations

  • Density of water is taken as 1000 g dm-3. Therefore, in a 100 mL aqueous solution of PbCl2, total mass of that solution is 100 g.

Step 1: First, we are going to find, maximum dissolving mass of PbCl2 in one liter (1 dm3) of water.

  • Maximum mass of dissolving PbCl2 in one liter (1dm3) = 0.99 * (1000/100) g
  • Maximum mass of dissolving PbCl2 in one liter (1dm3) = 9.9 g

Step 2: Now we know the maximum mass of dissolving (9.9 g) in 1 dm3 of water. Next step is, find the dissolved amount of PbCl2 moles in 1 dm3 of water.


Molecular mass of PbCl2 = 207.2 + (35.5*2)

Molecular mass of PbCl2 = 278.2 g mol-1


  • Maximum amount of dissolving PbCl2 in one liter (1 dm3) of water = 9.9 g / 278.2 g mol-1
  • Maximum amount of dissolving PbCl2 in one liter (1dm3) of water = 0.036 mol

Maximum concentration of PbCl2 can be dissolved in one liter (1dm3) of water = 0.036 mol dm-3



Step 3: Now, we are going to calculate solubility product of PbCl2 as following (solubility product equation for PbCl2).

Ksp, PbCl2 = [Pb2+(aq)] [Cl-(aq)]2



Step 4: Now, we need to find the concentrations of Pb2+ ion and Cl- ions in water.

PbCl2 poorly dissociate to Pb2+ ions and Cl- ions in water.

PbCl(aq) → Pb2+(aq) + 2Cl-(aq)


[Pb2+(aq)] = 0.036 mol dm-3

[Cl-(aq)] = 0.036 mol dm-3 * 2 = 0.072 mol dm-3


Ksp(PbCl2) = [Pb2+(aq)] * [Cl-(aq)]2

Ksp(PbCl2) = 0.036 mol dm-3 * (0.036 mol dm-3)2

Ksp(PbCl2) = 4.7 * 10-5 mol3 dm-9




Questions




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