This means that every reagent is at 10 times greater concentration than what you

Question

This means that every reagent is at 10 times greater concentration than what you need when you are actually using the reagent. E.g. You would need to do a 1:10 dilution of a 10 X buffer to get the buffer you want. If you are asked to make a very dilute solution of anything, consider making a more concentrated solution (something you can weigh) and then doing serial dilutions. Use the formulas in Labs 2 and 3 to help you figure out what final dilution you need and then figure out what individual dilutions you need to get you to the final dilution. Well-labeled tables help! Make 2 Liters of 1 X TE. Make 1 ml of a 0.25nM solution of NaOH, (FW: 40). You are given a maximum of 2 g of NaOH as your starting material. (Remember that you need to consider the equipment available in the lab). Make 500 ml of a 50 X solution of PBS. Use this (50X PBS Buffer) to make 1 liter of 0.5X PBS solution. Write out how you would do this and include the concentration of each ingredient in the 0.5X Buffer. Write out the calculations and provide instructions on how to prepare a 7.5 % (Na_2CO_3) in 0.75M NaOH. Remember NaOH is caustic so make sure to include appropriate precautions.

Explanation / Answer

Problem 1 :

1X TBE BUFFER 2 litres:

10mM tris

1mM = 121.4 g in 1 litre

10mM = 1214 g in 1 litre

I.e. 2428 g in 2 litres

1mM EDTA is made by adding 0.0186g EDTA into 40 ml distilled water and then adding NaOH to adjust the pH to 8

Problem 2:

0.25 nM solution of NaOH

1nM = 0.001 Molar

For 1 M NaOH = 40 g NaOH in 1 litre

1M NaOH = 0.04 g in 1 ml

For 0.001 Molar in 1 ml = 0.00004 g NaOH = 1nM

For 0.25 nM = 0.00001g NaOH in 1 ml

Problem 3 :

10mM Na2HPO4, 137mM NaCl, 2.7 mM KCl, 2mM KH2PO4, pH 7.4

1M values = Na2HPO4- 141.96g; NaCl = 58.44 g; KH2PO4 = 136.086 g; KCl = 74.5513 g

For 1x buffer 1000 ml - Na2HPO4=1.4196 g ; NaCl = 8.00628 g; KH2PO4 = 0.272172 g; KCl = 0.2012 g

1x buffer in 500 ml = Na2HPO4=0.7098 g ; NaCl = 4.00314 g; KH2PO4 = 0.1360 g; KCl = 0.1006 g

0.5x buffer in 500 ml = Na2HPO4= 0.3549 g ; NaCl = 2.00157 g; KH2PO4 = 0.068 g; KCl = 0.0503 g

0.5x buffer 1000 ml - Na2HPO4=0.7098 g ; NaCl = 4.00314 g; KH2PO4 = 0.1360 g; KCl = 0.1006 g

50x buffer in 500 ml - all the above values would be multiplied by 100

50x buffer in 500 ml = Na2HPO4= 35.49 g ; NaCl = 200.157 g; KH2PO4 = 6.8 g; KCl = 5.03 g

Problem 4:

7.5% Na2CO3 in 0.75M NaOH

Prepare 100 ml solution of 0.75M NaOH

1M = 4 g in 100 ml

0.75 M = 3 g in 100 ml , to this add 7.5 g of sodium bicarbonate as it is given in w/v percentage

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