Make 2 Liters of IX TE Make 1 ml of a 0.25 nM solution of NaOH. (FW:40). You are

Question

Make 2 Liters of IX TE Make 1 ml of a 0.25 nM 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 0.5 X 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 0.75 M NaOH. Remember NaOH is caustic so make sure to include appropriate precautions.

Explanation / Answer

Problem.1 :

A typical recipe for making 1X TE buffer is:

TE buffer is also called as T10E1 Buffer, and read as "T ten E one buffer".

To make a 100 ml solution of T10E1 Buffer, 1 ml of 1 M Tris-HCl (pH 8.0) and 0.2 ml EDTA (0.5 M) and made up with double distilled water up to 100ml. For making 2000 ml or 2L solution following calculations can be done:

Remember, we have to maintain the final concentrations of tris and EDTA as 10 mM (0.01M) and 1 mM (0.001M) in the 2L solution. Molarity equation is generally used for dilution where C1 and C2 are initial and final molarity while V1 and V2 are initial and final volumes.

C1V1 = C2V2

For Tris-HCl 1M x V1 = 0.01 x 2000 or V1 = 20 ml

For EDTA 0.5M x V1 = 0.001 x 2000 or V1 = 4ml

Volume of distilled water = 2000 -( 20+4) = 1976 ml

Therefore 2L of 1x TE buffer can be prepared by adding 20 ml of 1M tris , 4ml of 0.5M EDTA and 1976 ml of water.

Problem.2 :

Given 2 g NaOH is dissolved initially into 100ml distilled water.

Molarity of NaOH solution = W* 1000/M*V = 2 *1000/ 40 * 100 = 0.5 M

C1V1= C2V2

0.5 x V1 = 0.0005 x 100 or V1= 0.1 ml

Make up the 0.1 ml solution upto 100ml and resulting solution is 0.5mM NaOH.

2. 0.5 mM is diluted to 0.25 nM

0.5 x V1 = 0.000025 x 1000 or V1 = 0.05 ml

Make up 0.05ml of this solution upto 1000ml and resulting solution is 0.25nM NaOH.

Problem.3 :

Wt of NaCl = 8g (required for 1xPBS) x 50 = 400 g

Wt of Na2HPO4 = 1.44g (required for 1xPBS) X 50 = 72 g

Wt of KCl = 0.2g (required for 1xPBS) x 50= 10 g

Wt of KH2PO4 = 0.24g (required for 1xPBS) x 50 = 12g

Use 10 ml of 50x solution and dilute it to 1000ml with distilled water to get 1L of 0.5x solution.

1x PBS 50x PBS 0.5x PBS 137mM NaCl 6850 mM or 6.85M NaCl 68.5mM NaCl 10mM Na2HPO4 500mM or 0.5M Na2HPO4 5mM Na2HPO4 2.7mM KCl 135mM or 0.135M KCl 1.35mM KCl 2mM KH2PO4 100 mM or 0.1M KH2PO4 1mM KH2PO4

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