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This is a simple question but it has been so long since I have done any chemistry. Basically I would like to keep the concentration of A constant and the concentration of B variable and vice versa.

Assuming we want to make a 50 ml solution with constant concentration of A and variable concentration of B.From my knowledge I would use C1V1=C2V2 in order to dilute A to its fixed concentration lets say from 4% to 3%. But how do i go about keeping it constant and B concentration variable for a final 50 ml solution.

EDIT: Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?

2nd EDIT: Apologies, this is my first time using this site and was told to edit my question and add my comment. I didn't put values as I believed it was clear enough but I guess not.

@ y chung that is what I have carried out in order to create the solutions, so I'm guessing I did it correctly then.

@ bon thanks for the comment. I will tell you exactly what Im trying to accomplish. I have two solutions one is A with a concentration of 9% and one is B with a concentration of 7%. I want to create 5 solutions each 50 ml with a constant 5% concentration of A and varying concentration of B from 1-5 %.

What I have done is as follows: I reduced my A concentration to the required 5% by C1V1=C2V2 for a 50 ml solution getting 28 ml. I put 28 ml in each volumetric flask of 50 ml capacity.

I then calculated how much volume of B with the 7% concentration I would need to get it to say the 15 concentration based on my remaining volume of 22 (50-28). For the 1 % concentration this yielded a volume of B at 7% of 3.14 ml. I added that to my flask contain the 28 ml of A and made up the solution to 50ml.

The question is, is this correct? from what I gathered from y chung ,it is. Thanks

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    $\begingroup$ Could you please write the main question in full? You've only vaguely shaped your thoughts around it but since we haven't seen it we can't really tell what's being asked here. $\endgroup$ – M.A.R. Aug 11 '15 at 21:01
  • $\begingroup$ Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations? $\endgroup$ – John Aug 11 '15 at 21:05
  • $\begingroup$ @John - If you can edit your question to include your comments, that will be a big help in getting a useful response. $\endgroup$ – Todd Minehardt Aug 11 '15 at 22:44
  • $\begingroup$ Diluting A by a factor of four i.e. a 3:1 ratio, will give you a 1% solution. If you can use another diluent that is not A or B, this is a trivial exercise. Just add five different volumes of B (that are less than 12.5ml) to your five lots of 37.5ml A solution and top up to 50 ml. Unless I'm missing the point? $\endgroup$ – Beerhunter Aug 12 '15 at 17:09
  • $\begingroup$ @John You might have accidentally created two accounts here, which can be merged, so that there is only one account that you have to take care of. $\endgroup$ – Martin - マーチン Aug 14 '15 at 6:47
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Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.

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    $\begingroup$ Perhaps you could be more specific about exactly what you would do. At the moment this would probably be considered 'low quality'. $\endgroup$ – bon Aug 12 '15 at 10:20
  • $\begingroup$ He did not ask for what different concentrations of B he wanted. Also, He has done the keeping constant A part in his "EDIT" part himself. I don't think a very long and detail answer is required $\endgroup$ – y chung Aug 12 '15 at 10:26
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    $\begingroup$ Personally I have difficulty understanding what your answer means. I'm not asking for a long answer, just some additional explanation as to exactly what you are trying to say. $\endgroup$ – bon Aug 12 '15 at 10:34

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