[nickcrowley]

Hello.

I was preparing PCR solutions, mainly the MgCl₂ solution.  Accidentally (brain fart) I did not dilute correctly the 1M MgCl2 solution correctly to a final concentration of 25mM (millimolar).  However, I made the solution at 0.25mM.  This 0.25mM solution was already placed into PCR mastermix and cannot be removed.

So I must figure out how much 1M (at minimal volume) to add to increase the concentration to 25mM MgCl2.  I have tried the v1c1=v2c2 equation and came to 0.625uL (microliter) of 1M MgCl2 to add to acheive 25mM.  The PCR reaction total volume is 25uL.  Am I doing this correctly or am I completely wrong.

[allanf]

I seem to think it would be more complicated than v1c1 = v2c2, shouldn't you end up with a system of two unknowns?

Ok so the final volume and concentration should be 25uL and 25mM, and you need to achieve that by adding some unknown amount v1 of 1M to some amount v2 of 0.25mM right?

625 = v₁ + 0.25*v₂

As far as I can tell, unless you know v₂, how much of the 0.25mM solution in there, you probably can't solve for v₁, the volume of 1M solution you should add.

[nickcrowley]

we added 480uL of the 0.25mM solution.  It was supposed to be 480uL of 25mM solution.

[enahs]

(   (0.48mL + X) * 25mM - (0.48*0.25)    )/ 1000mM =x,

X= 0.01218461538 mL of 1000mM (1M)

Back check.

1000mM * 0.01218461538 mL = 12.18561538 mMoles
You already have 0.48mL, you are adding another 0.01218461538 mL, for a total of 0.4921856154mL

12.18561538 mMoles / 0.4928156154 mL = 25 mM



Maybe I am confusing what question you are asking?