CHEM 1111 Lecture Notes - Lecture 16: Spectrophotometry, Cuvette, Potassium Thiocyanate

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8 Nov 2014

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5 = 7. 3 e 05 m x . 005 l = 3. 65e 7 mol: initial moles of fe3+ 5 = (. 002 m) x (. 005 l) = 1 x 10-5 mol. 2 = (. 002 m) x (. 002 l) = 4 x 10-5 mol. 3 = (. 002 m) x (. 003 l) = 6 x 10-6 mol. 5 = (. 002 m) x (. 005 l) = 1 x 10-5 mol: equilibrium concentration of fe3+ 2 = ((4 x 10-5 mol) (5. 68e 8 mol)) / (. 01l) = ~ 0. 004 m. 3 = ((6 x 10-6 mol) (1. 404e 7 mol)) / (. 01l) = ~ 0. 0058 m. 5 = ((1 x 10-5 mol) (3. 99e 7 mol)) / (. 01l) = ~ 0. 0096 m. 3 = ((6 x 10-6 mol) (1. 152e 7 mol)) / (. 01l) = ~ 0. 0058 m. 5 = ((1 x 10-5 mol) (3. 65e 7 mol)) / (. 01l) = ~ 0. 0095 m: value of kc: Kc = [c]c[d]d / [a]a[b]b = [fescn][h] / [fe3+][hscn]

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Might you show me all steps in orden to understand everything that you did? Thanks.

Please I need help on numbers 7-10 !!!!! Will thumbs up and give great review !!

Hello, I am doing homework and I am not sure if my answers are correct, can someone help me verify if I am doing this right or wrong? My responses are underneath the questions.

You are given a protein solution with a concentration of 0.25 mg/ml. Answer these questions.

A.) We need 20 Âµg for an experiment. What volume of the protein solution do we need?

Convert 20 Î¼g and you will get 0.02 mg C = M/V rearrange for Volume V= M/C , V= (0.02 mg)/(0.25 mg/ ml) = 0.08 ml or 80 Î¼l

B.) Suppose we want to prepare a solution containing 150 Âµg of the protein at a concentration of 0.50 mg/ml. To do this we will first dry down enough protein solution to get 150 Âµg. How much solution do we need to start with? How much H2O do we add to get the desired concentration?

You will convert 150 Î¼g to 0.15 mg and divide by 0.25 mg/ ml to get 0.60 ml that needs to be dried down to obtain 0.5 mg/ml, then you will add enough water so that the total volume is 0.5 ml.

C.) If the protein has a M.W. = 15,000, express its initial concentration in moles/liter, in Âµmoles/ml and in Âµmoles/Âµl. If we want 2 Âµmoles for a reaction, what volume do we need?

Convert 0.25 mg/mL = 0.00025 g/mL = 0.00025 g/0.001 L = 0.25 g/L If M.W. of the protein is 15,000 g/mol then 0.25 g = 0.25/15,000 moles = 1.67 x 10^-5 moles. This many moles is contained in a liter (L) therefore concentration is 1.67 x 10^-5 moles/L = 1.67Î¼moles/L = 1.67 Î¼moles/1000 mL = 1.67 x 10^-3 Î¼moles/mL = 1.67 x 10^-3 Î¼moles/1000 Î¼L = 1.67 x 10^-6 Î¼moles/Î¼L. The concentration of the solution is 1.67 x 10^-6 Î¼moles/Î¼L, 1 Î¼mole is present in 1/(1.67 x 10^-6) Î¼L and if 2 Î¼moles are present in 2/(1.67 x 10^-5) Î¼L = 1.20 x 10^-5 Î¼L = 1.20 L

D.) Suppose we want 1 ml of a 10 Âµg/ml solution. How much H2O and protein stock must we add to get this?

C1V1 = C2V2 (250 Î¼g/ Î¼l)(V1) = (10 Âµg/ml)(1ml) V1= 0.04 ml or 40 Î¼l (Volume of protein stock for solution) Volume of water 1.0 ml- 0.04ml = 0.96 ml

E.) Suppose we want 100 Âµl of a 0.1 Âµg/Âµl solution. How much H2O and protein stock must we add to get this?

C1V1 = C2V2 (0.25 Î¼g/ Î¼l)(V1)= (0.1 Î¼g/ Î¼l)(100 Î¼l) V1=40.0 Î¼l (Volume of protein stock for solution) Volume of water 100 Î¼l- 40.0 Î¼l = 60.0 Î¼l

CHEM 1111 Lecture Notes - Lecture 16: Spectrophotometry, Cuvette, Potassium Thiocyanate

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