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I have two different mixtures:
1 gallon of "A mixture", which is 99% substance A and 1% water
An arbitrarily large supply of "B mixture", which consists of 4% substance B and 96% water.

How do I go about calculating how much of "B mixture" (plus extra water if needed) to add to the 1 gallon of "A mixture", to produce a final combined mixture consisting of 70% substance A, 2% substance B and the remaining 28% water?

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1 Answer 1

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The problem is that to get 2% (v/v) of B from a 4 % (v/v) solution of B then to make 1 litre of the final solution we would need 500 ml (0.5 L) of the 4 % stock solution of B (at 4 %).

As the stock of A is at 99 % (v/v) then when the concentration of B is 2 % then the highest concetration of A which is possible will be 49.5 %. You stated that you wanted to have a final mixture which is 70 % A and 2 % B.

I think that you would be better off getting a stronger stock solution of B, for example if you had a 20 % solution of B, then to make 1 L of a 2 % solution then you would need 100 ml of the stronger B solution.

Then if you were to have a 70 % final concetration of A then you would need 707 ml of the 99 % A solution. Then you would need to add 193 ml of diliuent (I assume water) to make the final mixture up.

I am making the assumption that the volume of the final mixture is equal to the sum of the volumes of the original liquids before mixing. It is interesting that ethanol / water mixtures have a volume which is rather less than the sum of the volume of pure ethanol and pure water.

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