Posted by Anonymous on Wednesday, August 7, 2013 at 10:06pm.
A stock standard contains 140 Eq/L of sodium and 5 Eq/L of potassium. How would the working standard of 140 m Eq/L and 5 m Eq/L be prepared?

Chemistry/lab math  DrBob222, Wednesday, August 7, 2013 at 10:17pm
The only thing you're doing is changing from equivalents/L to milliequivalents/L. Just dilute 1000 times; i.e., take 1 mL of the stock, add to a 1000 mL volumetric flask, dilute to the mark and mix thoroughly.
You can also use the c1v1 = c2vb2 I showed you how to use earlier.
140*x = 0.140*1000
x = 1 mL.
The problem didn't specify a volume; therefore, I chose 1000 mL as a convenience. Using any other volume works the same way. 
Chemistry/lab math  Anonymous, Wednesday, August 7, 2013 at 10:20pm
Thank you again!

Chemistry/lab math  Anon, Wednesday, August 7, 2013 at 10:32pm
Okay when you set up the equation likewise for potassium and solve it the same way is the answer supposed to be the same? I'm just confused as to whether the question is asking if each should be prepared differently or it's just one solution you're working with.

Chemistry/lab math  DrBob222, Wednesday, August 7, 2013 at 11:08pm
You're working with one solution containing two cations of different concentrations and the question is asking you how much of the stock solution to take to make a single solution containing BOTH cations of a different concentration. The procedure I showed you is for either one. Fortunately, both Na and K must be diluted by 1000 to achieve the concentration needed. You couldn't do it if you wanted different dilutions (well you could but you would need to make one dilution for Na and another for K). In this case, however, Na must be diluted by 1000 and K must be diluted by 1000 so the 1 dilution of 1:1000 will get it for both Na and K.
You can work it out it you wish for each and see that it is the same.
140*x = 0.140*1000
x = 1 mL. Add to 1L volumetric flask etc etc.
5*x = 0.005*1000
x = 1 mL. Add to 1L volumetric flask etc etc.
So the one dilution makes the solution containing both at the new concentration. You can see that any other set of numbers that don't have the same dilution factor just won't work. The person making up the problem took that into account. Bascally you worked a single problem and solved, actually, two problems. :). 
Chemistry/lab math  Anon, Wednesday, August 7, 2013 at 11:36pm
Thank you very much for this further clarification; I understand what it is asking now.