Question

At 20^oC, a saturated solution of silver acetate, AgCH₃CO₂, contains 1.0 g of the compound dissolved in 100. mL of solution. Calculate K_sp for silver acetate (ignoring hydrolysis)

Answer 1

Sol:-

Here we have to first calculate the solubility (S) of CH₃COOAg by using the formula :

Solubility (S) = Number of moles / Volume of solution in litre .

Solubility (S) is define as the "Number of moles of sparingly soluble salt dissolved in 1 litre of saturated solution" .

units of solubility (S) = mol/L

also number of moles of substance = given mass of substance in g / gram molar mass of substance .

therefore number of moles of CH₃COOAg = 1.0 g / 167 g/mol   ,( gram molar mass of CH₃COOAg = 167 g/mol )

                                                                  = 0.006 mol ( approx.)

therefore

Solubility (S) of CH₃COOAg = 0.006 mol / 0.100 L      ( because volume = 100 ml = 0.100 L)

                               = 0.06 mol/L

therefore solubility (S) of CH₃COOAg = 0.06 mol/L

Now the partial dissociation of CH₃COOAg ( Sparingly soluble salt ) in aqueous medium is :

CH₃COOAg(s) <-------------> CH₃COO^- (aq)   +   Ag⁺ (aq) .

                                              S mol/L                   S mol/L

Now the Expression of Solubility Product i.e K_sp for CH₃COOAg

K_sp = [CH₃COO^-] [ Ag⁺]

K_sp = S x S

K_sp = S²

K_sp = ( 0.06 )²                         ( because S = 0.06 mol/L)

K_sp = 0.0036

K_sp = 3.6 x 10^-3

Hence Solubility Product of CH₃COOAg = 3.6 x 10^-3

Note :- Solubility Product is the "product of molar concentration of products in which stoichiometric coefficient with respect to each of the product is raised to the power" .