Question

10.0mL of a 0.0500M KCl (aq) is titrated with 8.60mL of silver acetate (AgCH3COO (s) ). The silver acetate is in a saturated solution that contains 0.100M silver nitrate ( AgNO3). what is the Ksp and Molar solubility of the silver acetate (AgCH3COO)?​

Answer 1

Answer:

15 cm³ of liquid 'X' and 20cm³ of liquid 'y' are mixed at 20⁰C and the volume solution was measured to be 35.1cm³

Answer 2

Given :

Molarity of KCl (M₁) = 0.05M = 5*10⁻²M

Volume of KCl taken (V₁) = 10mL = 10*10⁻³L

Volume of CH₃COOAg (V₂) = 8.6mL = 8.6*10⁻³L

To find :

Ksp and Molar solubility of silver acetate (AgCH₃COO)

Solution :

1) First find the molarity(M₂) of CH₃COOAg by titration.

M₁V₁ = M₂V₂

M₂ = M₁V₁

        V₂

M₂ = 5*10⁻² * 10*10⁻³

            8.6*10⁻³

M₂ = 5.8139*10⁻²M

Molarity of CH₃COOAg (M₂) = 0.058139M

Molarity and Molar solubility are same quantity, having same wnit i.e., mol/L

∴ Molar solubility (S) of CH₃COOAg = 0.058139M

2) Now, find solubility of silver acetate

Molar solubility (mol/L) =   Solubility in g/L    

                                         Molar mass in g/mol

Solubility = Molar solubility * Molar mass

Solubility = 5.8139*10⁻² * 199.91

Solubility = 970.3*10⁻²

Solubility = 9.703g/L

3) Ksp of silver acetate

AgNO₃ → Ag⁺ + NO₃⁻          (Completely dissociated salt)

AgCH₃COO ⇄ Ag⁺ + CH3COO⁻          (Sparingly soluble salt)

Due to common ion effect equilibrium shift to left hand side and their is more amount of undissolved CH₃COOAg in solution.

∴ Ksp of silver acetate = [Ag⁺]¹ [CH₃COO⁻]¹

[Ag⁺]¹ = 1 * S mol/L

[CH₃COO⁻]¹ = 1 * S mol/L

S is molar solubility

∴ Ksp = S * S

Ksp = S²

Ksp = (5.8139 * 10⁻²)²

Ksp = 33.801*10⁻⁴

Ksp = 3.3801*10⁻³

Therefore, Molar solubility is 5.8139*10⁻² and Ksp is 3.3801*10⁻³.