Question

Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 3.462 using the systematic treatment of equilibrium. Ksp(Ag2CO3) = 8.46 × 10–12; Ka1(H2CO3) = 4.45 × 10–7; Ka2(H2CO3) = 4.69 × 10–11.

Answer 1

Solubility equilibrium: Ag,CO3(s) = 245+ (aq) + Co" (aq) Solubility product of Ag,CO3, Kg = = 8.46x 10-12 pH = 3.462 pOH = 14.00 -3.462=10.538 [OH-] = 10-10.38 = 2.90x10-1 M - - - - - - - - - Desired equilibrium: Ag, C (s) + H2O(l) 2 2 Ag+ (aq) + HCO3(aq)+ OH- Solubility product, Km = [18+] [HC05][0H-] ={[Ag*][coß-]] ([H*TOH-]}x_{HCO3] [Co-][H+] = [[A8+][c03]]>([*][oH])* TC03- | H+] [HCO] = K, ⓇK,* **K (H,CO) = ( 8.46x10-la) x (1.00x10+4)*7 693 10-11 = 1.80x10-25 1 - - - - - - - - - - ICE table: Ag, Co, (s) + H20(1) 2 Initial (M) Change (M) Equilibrium (M): 2 Ag+ (aq) + HCO3(aq)+ OH- 0 2.90x10-11 +s +s 2.90x10-11+s + Solubility product, Ky=[Ag*)[;][ OH-] 1.80x10-15 = (25) x(s) (2.90x10-+s) s = 1.46x10*M Thus, the molar solubility of Ag, CO, is 1.46x10*M