Unformatted text preview: y = 0.446 x = 2.50 5 Solving for the crossing point by using the 2 best fit equations gives, c = 3.76
Q 5 Al 10 M. +10 pts for
Q 4 graph and analysis (a) Since c = 3.7 10 M and complex formation is saturated when c = 3.76 10
5 2+ M, the complex must be 1:1, or AlQ . (b) AlQ 2+ = (0.500) / (3.7 10 ) = 1.4 10 5 + 10 pts for concentration determination. The strategy here uses data in the saturation region (flat) to determine the formation complex where you know that all of the limiting reagent is complexed (b). From the spreadsheet, m = 0.02613 and b = 0.2125 (c). From LINEST, s = 0.000829, s = 0.01254
m b –6 (d) From the spreadsheet, c (e) sc = cx Pd(II) = 8.13
6 10 M sm m 2 + sb b 2 = 8.13 10 0.000829 0.026130 2 + 0.01254 0.2125 2 = 5.45 10 7 M First, use the data where [A]/[Cu2+] > 10 where you know that the total absorbance is proportional to the [Cu2+]total
–4 3 = 0.759/2.15 10 = 3.53 10
–4 Now, use the other information to calculate the formation constant [CuA22] = 0.654/3.53
2+ –4 103= 1.8526
–4 10 M
–5 [Cu ] = 2.15 10 – 1.8526 10 = 2.974 10 M 2– –4 –4 –5 [A ] = 4.00 10 – 2 1.8526 10 = 2.9486 10 M f =
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 Spring '08
 BAILEY
 pH, Reagent, saturation, best fit equations

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