Analytical Chem istry - DePauw University

262 Analytical Chemistry 2.0From this point, we made two assumptions, simplifying the problem toone that was easy to solve.[ HO ] = KC = ( 68 . × 10 )( 10 . ) = 26 . × 10+ −4 −23 a HFYou also can solve this set of simultaneousequations using Excel’s Solver function.To do so, create the spreadsheet in Figure6.18a, but omit all columns other thanA and B. Select Solver... from the Toolsmenu and define the problem by using B6for Set Target Cell and setting its desiredvalue to 0, and selecting B1 for By ChangingCells:. You may need to play with theSolver’s options to find a suitable solutionto the problem, and it is wise to try severaldifferent initial guesses.The Solver function works well for relativelysimple problems, such as findingthe pH of 1.0 M HF. As problems becomemore complex—such as solving an equilibriumproblem with lots of unknowns—the Solver function becomes less reliablein finding a solution.Although we did not note this at the time, without making assumptionsthe solution to our problem is a cubic equation+ +[ HO ] + K [ HO ]3 23 a 3+− ( KC + K )[ HO ] − K K = 0a HA w 3 a w6.64that we can solve using Excel’s Solver function. Of course, this assumes thatwe successfully complete the derivation!Another option is to use Excel to solve the equations simultaneouslyby iterating in on values for [HF], [F – ], [H 3 O + ], and [OH – ]. Figure 6.18ashows a spreadsheet for this purpose. The cells in the first row contain ourinitial guesses for the equilibrium pH. Using the ladder diagram in Figure6.14, choosing pH values between 1 and 3 seems reasonable. You can addadditional columns if you wish to include more pH values. The formulasin rows 2–5 use the definition of pH to calculate [H 3 O + ], K w to calculate[OH – ], the charge balance equation to calculate [F – ], and K a to calculate[HF]. To evaluate the initial guesses, we use the mass balance expressionfor HF, rewriting it as−−[ HF] + [ F ] − C = [ HF] + [ F ] − 10 . = 0HFand entering it in the last row. This cell gives the calculation’s errorFigure 6.18b shows the actual values for the spreadsheet in Figure 6.18a.The negative value in cells B6 and C6 means that the combined concentrationsof HF and F – are too small, and the positive value in cell D6 meansthat the combined concentrations are too large. The actual pH, therefore,must lie between 2.00 and 1.00. Using these pH values as new limits forthe spreadsheet’s first row, we continue to narrow the range for the actualpH. Figure 6.18c shows a final set of guesses, with the actual pH fallingbetween 1.59 and 1.58. Because the error for 1.59 is smaller than that for1.58, we will accept a pH of 1.59 as the answer. Note that this is an agreementwith our earlier result.Practice Exercise 6.14Using Excel, calculate the solubility of AgI in 0.10 M NH 3 without makingany assumptions. See our earlier treatment of this problem for therelevant equilibrium reactions and constants.Click here to review your answer to this exercise.