Sample calculations converting from t to a for

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Sample Calculations Converting from %T to A: For solution #1: %T = 58.1% A = 2 - log(%T)
A = 2 - log(58.1) A = 0.236 Determining the Concentration of Fe 3+ in Dilute Iron Solution: M 1 V 1 =M 2 V 2 M 1 = 2.50 x 10 -3 M V 1 = 5.00 mL (2.50 x 10 -3 M) (5.00 mL) = M 2 (100.00 mL) V 2 = 100.00 mL M 2 = 1.25 x 10 -4 M Dilution Calculations: For solution #1 M 1 V 1 =M 2 V 2 M 1 = 1.25 x 10 -4 M V 1 = 5.00 mL V 2 = 25 mL (1.25 x 10 -4 M) (5.00 mL) = M 2 (25 mL) M 2 = 2.5 x 10 -5 M Determining ɛ 480 from Beer-Lambert Plot: y = mx + b y = 8920x - 0.0133 m = 8920
ɛ 480 = 8.92 x 10 3 M -1 cm -1 Determination of [Fe(SCN) 2+ ] eqm For solution #4: A 480 = εl [Fe(SCN) 2+ ] eqm A 480 = 0.785 ε 480 = 8.92 x 10 3 M -1 cm -1 l = 1.00 cm [Fe(SCN) 2+ ] eqm =A/Ɛl [Fe(SCN) 2+ ] eqm = 0.785/ 8.92 x 10 3 ? −1 𝑐? −1 1.00 [Fe(SCN) 2+ ] eqm = 8.80 x 10 -5 M Determination of the initial concentrations of Fe3+ and SCN-: For solution #4 for [Fe 3+ ] 0 : M 1 V 1 =M 2 V 2 (2.50 x 10 -3 M) (10.00 mL) = M 2 (25 mL) M 2 = 1.0 x 10 -3 M Determination of the [Fe 3+ ] concentration: For solution #4: [Fe 3+ ] eqm = [Fe 3+ ] 0 - [Fe(SCN) 2+ ] eqm [Fe 3+ ] 0 = 1.0 x 10-3 M [Fe(SCN) 2+ ] eqm = 8.80 x 10-5 M [Fe 3+ ] eqm = (1.0 x 10-3 M) – (8.80 x 10-5 M)
[Fe 3+ ] eqm = 9.12 x 10-4 M Determination of the [SCN-] concentration: For solution #4: [SCN - ] eqm = [SCN - ] 0 - [Fe(SCN) 2+ ] eqm [SCN - ] 0 = 1.0 x 10 -3 M [Fe(SCN) 2+ ] eqm = 8.80 x 10 -5 M [SCN - ] eqm = (1.0 x 10 -3 M) – (8.80 x 10

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