Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry281Concentrations Hg 2 Cl 2 (s) Hg 22+ (aq) + 2Cl – (aq)Initial solid 0 0.10Change solid +x +2xEquilibrium solid x 0.10 + 2xand substituting into the thermodynamic solubility product for Hg 2 Cl 2 ,leave us with22+ 2 − 2K = a a = γ [ Hg ] γ [ Cl ] = 1.2× 10 −18sp 2+ − 2+ −Hg2Cl Hg 22Cl12 . × 10 − = ( 0. 351)( x)( 075 . )( 0. 10 + x)18 2 2Because the value of x is likely to be small, let’s simplify this equation to12 . × 10 − = ( 0. 351)( x)( 075 . )( 0. 10)18 2 2Solving for x gives its value as 6.1 × 10 –16 . Because x is the concentrationof Hg 2 2+ and 2x is the concentration of Cl – , our decision to ignore theircontributions to the ionic strength is reasonable. The molar solubility ofHg 2 Cl 2 in 0.10 M NaCl is 6.1 × 10 –16 mol/L. In Practice Exercise 6.8,where we ignored ionic strength, we determined that the molar solubilityof Hg 2 Cl 2 is 1.2 × 10 –16 mol/L, a result that is 5× smaller than the itsactual value.Click here to return to the chapter.Practice Exercise 6.14To solve this problem, let’s set up the following spreadsheetA1 pI = 32 [I-] = = 10^–b13 [Ag+] = = 8.3e–17/b24 [Ag(NH3)2+] = = b2 – b35 [NH3] = = (b4/(b3*1.7e7))^0.56 [NH4+] = = 0.10 – b5-2*b47 [OH-] = = 1.75e–5*b5/b68 [H3O+] = = 1.00e–14/b79 error = b3 + b4 + b6 + b8 – b2 – b7copying the contents of cells B1-B9 into several additional columns. Seeour earlier treatment of this problem for the relevant equilibrium reactionsand equilibrium constants. The initial guess for pI in cell B1 givesthe concentration of I – in cell B2. Cells B3–B8 calculate the remainingconcentrations, using the K sp to obtain [Ag + ], using the mass balance oniodide and silver to obtain [Ag(NH 3 ) 2 + ], using b 2 to calculate [NH 3 ], usingthe mass balance on ammonia to find [NH 4 + ], using K b to calculateB