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Making TrisHCL containing MgCl2


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#1 agp23

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Posted 15 April 2009 - 02:12 PM

How can I make a 100mM solution of TrisHCL containing 1 mM MgCl2 in a total volume of 100 mls ? This is for my science project, my teacher gave me a bottle of trizma HCL and 1 M stock solution of MgCl2( 100 ml). This is what I am doing. Pls let me know if this is correct. I added 15.67 g of TrisHCL in 100 ml and then I added 100 microliters of MgCl2 from the stock solution to get a 1 mM concentration. Am I doing this right ? Please help.

#2 bob1

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Posted 15 April 2009 - 04:30 PM

Trizma HCl (Tris HCl) has a molecular mass of 157.6 g/mole. given n=v*c (in moles and litres) n= 0.1 *0.1 = 0.01 moles required for 100 ml of 10 mM solution. Therefore given mass=n* molar mass = 0.01*157.6 = 1.576 g

For dilutions: Vi*Ci=Vf*Cf where V and C are volume and concentration respectively, and i and f are initial and final respectively. You have final volume (100 ml) and final concentration (1 mM) and initial concentration (1 M = 1000 mM - conserve units, it won't work otherwise). rearrange equation to find Vi gives you Vi=(Vf*Cf)/Ci = (100 * 1)/1000 = 0.1 ml (100 microlitres).

Dissolve the tris HCl in 80 ml water, check pH if necessary, add 100 ul of MgCl2 and tip the whole lot into a 100 ml volumetric flask, rinse the initial container into the flask as well, to ensure that you have all the reagents in the flask, and top up to 100 ml.

Note: Most people when making tris (Trizma =brand name tris from Sigma-Aldrich) solutions will use plain tris-base and add HCl to pH it to the desired pH, the result is the same, but it is much cheaper.




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