How To Calculate Solubility

How To Calculate Solubility

Example 1

30cm 3 of an insoluble solution of salt NaX was dissolved in water at 25°C and evaporated until dryness, and only 2.5g of remaining solid material was left. Determine the amount of salt solubilized in the water 25°C as moles per dm 3.(Na = 23, X = 19,)

Solution

30cm 3 of saturated solution has 2.5g

1,000cm 3. in the mixture will be able to contain

1000cm330cm3x2.5g1=83.3g/dm3

Solubility in mols/dm3 = Solubilitying/dm3MolarMass

Molar mass Nax = 19 + 23 = 42

= 83.342 = 1.98mol/dm3

Or

Solubility = MassofsoluteVolumeofsolventx1000MolarMass

= 2.530×100042

= 1.98mol/dm3

Example 2

At 25°C, 23g of KCl dissolves into 100 centimeters 3.0 of water, resulting in an aqueous solution. What is the salt’s solubility in 1. Grammes for every dm 3. of the solvent at this temperature?

  1. Moles per dm3 [K=39, Cl=35.5]

Solution

(1) Solubility in g/dm3 = MassofsoluteVolumeofsolventx10001

= 23g100cm3x1000cm31

= 230g/dm3

(2) RMM of KCl = 39+35.5 = 74.5

Solubility in mol/dm3 = Solubilityingdm-3Molarmass

= 23074.5

= 3.09mol/dm3

Example 3

An evaporating dish weighed 31.18g, when filled with saturated solution of copper teteraoxosulphate(vi) solution at 32degC it weighed 48.39g, after evaporation to dryness the evaporating dish and residue weighed 33.98g.

Determine the concentration of solubility for copper (II) Tetraoxosulphate (IV) crystals at temperatures of 32 degrees Celsius in grams in 100g of the solvent.

Solution

At 32degc

Mass of the dish that evaporates + CuSO 4 solid33.98g
Mass of dish that evaporates31.18g
Volume of CuSO 4(s) solid=2.80g
Mass of the dish used to evaporate + CuSO 4(aq) solution48.39g
Mass of dish for evaporation + CuSO 4(s) solid33.98g
The amount of solvent=14.41g
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14.41g of the solvent were saturated with 2.80g of CuSO 4(s) solid

1g solvent is saturated with 2.8014.41 of CuSO 4(s)solid

The solvent contained in 100g was saturated with

2.8014.41x1001ofCuSO4(s)

= 19.43g/100g of solvent

Example 4

250cm 3 of saturated solution of Sodium Chloride has 5.85g of salt that has been that is dissolved within it at the temperature of room. How much is the solubility salt? (i) grams per dm 3. (ii) mole/dm 3 [Na = 23 Cl =35.5]

Solubility in g/dm 3 =

Massofsolutevol/massofsolventx10001

= 5.85g250cm3x10001

= 23.4g/dm3

Solubility in mol/dm3 = SolubilityingMolarMass

Molar mass NaCl is 23 + 35.5 = 58.5

= 23.458.5

= 0.4mol/dm3

 

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