Solubility Concept Explanation

We are given the solubility product constant K_sp for Ag₂CrO₄ as $1.1\times {10}^{-12}$ at 298 K. We need to find its solubility in a 0.1 M AgNO₃ solution.

Step 1: Write the Dissociation Equation

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)

Step 2: Write the K_sp Expression

K_sp = [Ag⁺]²[CrO₄²⁻] = $1.1\times {10}^{-12}$

Step 3: Account for Common Ion Effect

The solution already contains Ag⁺ ions from 0.1 M AgNO₃. Let the solubility of Ag₂CrO₄ be S mol/L. Then:

[Ag⁺] = 0.1 + 2S ≈ 0.1 M (since S is very small)

[CrO₄²⁻] = S

Step 4: Substitute into K_sp Expression

K_sp = (0.1)² × S = $1.1\times {10}^{-12}$

0.01 × S = $1.1\times {10}^{-12}$

Step 5: Solve for S

S = $\frac{1.1\times {10}^{-12}}{0.01}$ = $1.1\times {10}^{-10}$ mol/L

Final Answer

The solubility of Ag₂CrO₄ in 0.1 M AgNO₃ solution is $1.1\times {10}^{-10}$ mol/L

Related Topics

• Solubility Product Constant (K_sp)
• Common Ion Effect
• Ionic Equilibrium
• Precipitation Reactions

Important Formulae

For a salt A_xB_y ⇌ xA^y+ + yB^x−:

K_sp = [A^y+]^x[B^x−]^y

Common Ion Effect: Presence of common ion decreases solubility

Solubility (S) in pure water: S = $\sqrt[x+y]{\frac{K_{\text{sp}}}{x^x{y}^y}}$