Results:
Part 1
The λmax=x when in the first derivative. y = -0.0002x2 + 0.1626x - 30.972 y’=(2)-0.0002x+0.1626+C 0=-0.0004x+0.1626
-0.1626=-0.0004x
-0.1626/(-0.0004)=x
X=406.5=λmax
Part 2: pH Absorbance
Blank
0
5
0
6
0.033
7
0.23
7.5
0.293
8
0.438
8.5
0.429
9
0.426
10
0.476
y = -0.0035x3 + 0.0637x2 - 0.2406x + 0.0014 y = -0.0035x3 + 0.0637x2 - 0.2406x + 0.0014 y’= (3)-0.0035x2 + (2)0.0637x - 0.2406 +C y’=-0.0105x2+0.1274x-0.02406 y”=(2) -0.0105x+0.1274+C y”=-0.021x+0.1274 0=-0.021x+0.1274
-0.1274=-0.021x
-0.1274/(-0.021)=x
X=6.06=pKa
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