Calculate the molar solubility of Ag₂ CO3 in a solution buffered to a pH of 9.00. (Ksp (Ag2CO3) = 8.1 × 10-12, K₁ (H2CO3) = 4.5 × 10-7, K2(H2CO3) = 4.7 × 10-11) Molar solubility = M Step 1. Write Equations for the Pertinent Equilibria Three equilibria need to be considered: Ag₂CO3(a) 2Ag* + CO₂² H2CO3 + H2O = HCO₂¯ + H₂O+ HCO3 + H2O CO₂² + H₂O+ Step 2. Define the Unknown Since 2 mol of Ag is formed for each mole of Ag₂ CO₂ dissolved, 1 2 S = Solubility = [Ag+] Step 3. Write All Equilibrium-Constant Expressions = 8.1 x 10-12 (1) K₁p = [Ag+]² [CO₂²] = [H3O+] [HCO3] [H2CO3] H_O+][CO*] [HCO3-] = K₁ = 4.5 x 10' (2) = K₁ =4.7 x 10-11 (3) Step 4. Write Mass-Balance Expressions [Ag+] = 2[CO2]+2[HCO3] + 2[H₂CO3] (4) Step 5. Write the Charge-Balance Expression No charge balance because an unknown buffer is maintaining the pH. Step 6. Count the Number of Independent Equations and Unknowns We have four unknowns ([Ag+], [CO2], [HCO3], and [H₂CO₂]). Step 7a. Make Approximations No approximations needed, because we have 4 equations and 4 unknowns. Step 8. Solve the Equations For pH = 5.10, [H₂O+] = 7.94 × 10-6 Substituting [H3O+] 7.94 × 100 M into equation (3) and rearranging gives - 7.94 × 10-6 [CO,³] [HCO3]= = 1.69 x 4.7 x 10-11 105 [CO,³] Substituting this relationship and [H3O+] = 7.94 x 10 M into equation (2) and rearranging gives [H2CO3] 7.94 × 10-6 × 1.60 × 10[CO,2] 4.5 x 10-7 - = 2.98 × 10 [CO*] Substituting these last two relationships into equation (4) gives - [Art]=2x [CO*]+2×1.69 × 10®[CO,*]+2×2.98 × 10 [CO ] = 6.30 × 10" Substituting this last relationship into equation (1) gives Kap= [Ag+]*[Ag+] 6.30 x 106 [Ag] = 8.1 × 10 - S=[A] 8.1 x 10-12 -12 x 6.30 x 10" 3.7 x 10-2 = 1.9 x 10-2 M 10° [CO₂2]