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?
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.
Thank you again!
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.
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. :-).
Thank you very much for this further clarification; I understand what it is asking now.
To prepare a working standard of 140 mEq/L of sodium and 5 mEq/L of potassium, we can start with the stock standard and make appropriate dilutions. Here's how you can do it:
1. Determine the required volume of the working standard that needs to be prepared. Let's say you want to prepare 1 liter of the working standard solution.
2. Calculate the amount of stock standard solution needed to achieve the desired concentration. Since the stock standard has a concentration of 140 Eq/L for sodium, you will require 1 liter × 140 mEq/L = 140 mEq of sodium. Similarly, for potassium, you will require 1 liter × 5 mEq/L = 5 mEq of potassium.
3. Dilute the stock standard with an appropriate volume of solvent to achieve the desired concentration. In this case, the desired concentration is already in mEq/L, so the dilution factor will be 1. You can simply use a 1:1 dilution by mixing equal volumes of the stock standard and the solvent. For example, if you want to prepare 1 liter of the working standard, take 500 mL of the stock standard and add 500 mL of the solvent.
4. Use a measuring cylinder or a pipette to accurately measure the specified volumes of the stock standard and the solvent.
5. Mix the stock standard and the solvent thoroughly to ensure homogeneity.
By following these steps, you can prepare a working standard solution with a concentration of 140 mEq/L of sodium and 5 mEq/L of potassium.