Analytical Chem istry - DePauw University

282 Analytical Chemistry 2.0[OH – ], and using K w to calculate [H 3 O + ]. The system’s charge balanceequation provides a means for determining the calculation’s error.+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] −[ I ] − [ OH ] =03 2 4 3The largest possible value for pI—corresponding to the smallest concentrationof I – and the lowest possible solubility—occurs for a simple, saturatedsolution of AgI. When [Ag + ] = [I – ], the concentration of iodide is[ I ] = K = 83 . × 10 = 91 . × 10− −17 −9spcorresponding to a pI of 8.04. Entering initial guesses for pI of 4, 5, 6, 7,and 8 shows that the error changes sign between a pI of 5 and 6. Continuingin this way to narrow down the range for pI, we find that theerror function is closest to zero at a pI of 5.42. The concentration of I – atequilibrium, and the molar solubility of AgI, is 3.8 × 10 –6 mol/L, whichagrees with our earlier solution to this problem.Click here to return to the chapterPractice Exercise 6.15To solve this problem, let’s use the following function> eval = function(pI){+ I =10^–pI+ Ag = 8.3e–17/I+ AgNH3 = Ag – I+ NH3 =(AgNH3/(1.7e7*Ag))^0.5+ NH4 =0.10-NH3 – 2*AgNH3+ OH =1.75e–5*NH3/NH4+ H3O =1e–14/OH+ error = Ag + AgNH3 + NH4 + H3O – OH – I+ output = data.frame(pI, error)+ print(output)+ }The function accepts an initial guess for pI and calculates the concentrationsof species in solution using the definition of pI to calculate [I – ],using the K sp to obtain [Ag + ], using the mass balance on iodide and silverto obtain [Ag(NH 3 ) 2 + ], using b 2 to calculate [NH 3 ], using the mass balanceon ammonia to find [NH 4 + ], using K b to calculate [OH – ], and usingK w to calculate [H 3 O + ]. The system’s charge balance equation provides ameans for determining the calculation’s error.+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] −[ I ] − [ OH ] =M03 2 4 3The largest possible value for pI—corresponding to the smallest concentrationof I – and the lowest possible solubility—occurs for a simple, saturatedsolution of AgI. When [Ag + ] = [I – ], the concentration of iodide is