Solubility product calculator software

Solubility Product

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Introduction

Some compounds dissolve in water as molecules while others, called electrolytes, dissociate and dissolve as charged species called ions. Compounds which exist as solid ionic crystals are mostly highly soluble in water. Ionic compounds dissolve to the point where the solution is saturated and no more solid can dissolve. The concentration of the saturated solution is termed the solubility of the substance. In some cases the solubility may be very high and a large amount of the solid may dissolve before the solution is saturated. The differency in ability to dissolve in water are large. Some highly soluble salts dissolve very easily (1/2 kg in 1 kg water). Other salts doesn't seem to dissolve at all (1*10^-15 g in 1 kg water), even if the temperature is the same. The description "highly soluble" generally means soluble to at least the extent of forming 0.1 to 1.0 molar aqueous solutions. Salts which are less soluble in water than this at room temperature are called slightly soluble salts.

Solubility Guideline

Soluble salts:
1) Salts of alkali metals (group I) are highly soluble. Exception is KClO₄ (moderately soluble)
2) Nitrates and ammonium salts.
3) Metal halides are generally highly soluble. Exceptions are those of : Pb⁺², Ag⁺ og Hg₂⁺².
4) Most sulfate salts. Exeptions are those of: Ca⁺², Ba⁺², Sr⁺², Pb⁺² and Hg₂⁺².
Insoluble salts:
1) Salts of carbonates, phosphates, hydroxides and sulfides are usually insoluble. Exeptions are alkali metals (group I) and following moderately soluble salts: Ca⁺², Ba⁺², Sr⁺².
2) Metal sulfides are generally insoluble.

Solubility Product Constant

Ionic compounds dissolve to the point where the solution is saturated and no more solid can dissolve. The concentration of the saturated solution is termed the solubility of the substance. In such a saturated solution, equilibrium is established. Ions will form solid in the same extent as solid dissociate and form ions. The solubility product constant or solubility product K_sp is a temperature-dependent constant referring to this stage.
If M_xA_y dissociate into (cations) M^+m and (anions) A^-a The expression for solubility product will be:
K_sp=[M^+m]^x [A^-a]^y

The Common Ion Effect

The concentrations of ions in solution are affected by all equilibria and all species present in the solution. The simplest and most significant such effect is called the common ion effect. The common ion effect is observed whenever an ion in solution is common to two different salts which serve as its sources. Addition of the second salt adds the common ion, which is a product of the dissolution of the first. The effect of adding the product ion will be to decrease the solubility of the first salt.
If a salt M1_xA1_y (e.g. BaF2 : x=1, y=2) is added into a solution already containing a common ion e.g. M2^+m (Ba⁺²), the expression for the solubility product will be:
K_sp=[M1^+m+M2^+m]^x [A1^-a]^y

Examples from program - problems and solutions

Ex.1:How many moles of AgCl will dissolve in 0.5 kg of water? K_sp(AgCl)=1.77E-10.
Solution:
1) Insert equation of dissociation: AgCl = Ag+ + Cl-
2) Insert 1.77E-10 in K_sp-field
3) Insert mass (solvent): 0.5 kg
4) Calculate

Ex.2: 1.07722E-17 moles of Ag₂S will dissolve in 1 kg of water. Calculate K_sp(Ag₂S)
Solution:
1) Insert equation of dissociation: Ag₂S = 2Ag+ + S-2
2) Insert 1.07722E-17 in Ag₂S-field
3) Insert mass (solvent): 1 kg
4) Calculate

Ex.3: How many grams and moles of BaSO₄ (K_sp=1.8E-10) will dissolve in 1000 kg of water?
Solution:
1) Insert equation of dissociation: BaSO₄ = Ba+2 + SO4-2
2) Insert 1.8E-10 in K_sp-field
3) Insert mass (solvent): 1000 kg
4) Calculate

Ex.4: How many grams of BaSO₄ (K_sp=1.8E-10) will dissolve in a 1000 kg 0.1M Na₂SO₄-solution ?
Solution:
1) Insert equation of dissociation: BaSO₄ = Ba+2 + SO4-2
2) Insert 1.8E-10 in K_sp-field
3) Insert Na₂SO₄-conc. (0.1) in SO₄^-2-common ion effect field.
4) Insert mass (solvent): 1000 kg
5) Calculate

Ex.5: How many grams of MgF₂ (K_sp=7.42E-11) will dissolve in 0.5 kg of water?.
Solution:
1) Insert equation of dissociation: MgF₂ = Mg+2 + 2F-
2) Insert 7.42E-11 in K_sp-field
3) Insert mass (solvent): 0.5 kg
4) Calculate

Ex.6: How many grams of MgF₂ (K_sp=7.42E-11) will dissolve in a 0.5 kg 0.1M NaF-solution?.
Solution:
1) Insert equation of dissociation: MgF₂ = Mg+2 + 2F-
2) Insert 7.42E-11 K_sp-field
3) Insert (0.1) in F^--conc. common ion field.
4) Insert mass (solvent): 0.5 kg
5) Calculate

Solutions 1-6

Ex.1: 6.652E-6 mol
Ex.2: 5E-51
Ex.3: 3.13 g , 0.013 mol
Ex.4: 0.00042 g
Ex.5: 0.00825 g
Ex.6: 2.31E-7 g

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