1) How would you make: show your work

A) 1% (w/v) CMC / 0.5% (v/v) Tween-80 in water, 100 mL

B) 2% (w/w) Vitamin E TPGS in water, 100 mL

C) 100 mM citric acid in water, 100 mL (MW = 192.124 g/mol)

D) 100 mM acetate buffer (pH ~5, pKa = 4.76), 1000 mL (in moles)

E) 20 mg/mL ascorbic acid, 15 mL

F) 1 M HCl from conc. HCl (~12.1 M), 100 mL

G) 0.85% NaCl in water from 0.9% NaCl in water, 100 mL

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Tod ThielLv2

30 Jan 2019

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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

You are to make 90.0 mL of a 0.27 M buffer at ph 7.20 using the acid form of the Good buffer EPPS (MW = 252.332 g / mol, pKa = 8.00)

Questions:

1. Using the molarity, volume, and molecular weight, calculate how many grams of solid EPPS you need to weigh out?

2. Using the Henderson-Hasselbalch equation and the information given, calculate the value of [A-] / [HA]

3. Using the value of [A-] / [HA] calculated in question 2 and the concentration given at the beginning (Conc.), solve for x in this equation:

[A-] / [HA] = x / (Conc. -x)

4. The unknown x is in units of Molar. Given the volume from the beginning, how many moles of NaOH would this be?

5. Calculate how many mL of a 1 M NaOH solution you would need for the number of moles calculated in the previous question.

When making a buffer using two solutions - one of the conjugate acid, one of the conjugate base - the calculations are very similar.

You are to make 100.0 mL of a 0.50 M acetate buffer ( pKa = 4.72) at ph 5.00 using 0.50 M stocks of acetic acid and sodium acetate

6. Using Henderson-Hasselbalch and the information given, calculate the value of [A-] / [HA]

7. Reuse the equation from questions 3, but in place of Conc. use the final volume (100 mL) of the solution. The unknown (x) is now the volume of base form we need to add. Using [A-] / [HA] calculated in questions 6 and the equation shown, solve for x.

8. If the final volume is the total of the acid and base solutions, how many mL of the acid form do you need to combine with the base form in the previous question?