I have a 250mls of 8.4% bicarb. solution and want a 5% solution of the same in 250mls.

Question

250mls of 8.4% solution to become 250 mls of a 5 % solution?

Answer 1

8.4% solution contains 0.084×250=21mL of pure bicarbonate.

5% solution contains 12.5mL, so we have to lose 8.5mL of bicarbonate.

Adding x mL of pure water will increase the dilution but will increase the quantity to 250+x mL.

In this amount of solution, 5% (1/20) has to be pure bicarbonate, so (250+x)/20=21. In other words we have the same amount of bicarbonate in a larger quantity of water. We can solve for x:

250+x=420, x=170mL. So, when we add 170mL of pure water to the 8.4% bicarbonate solution we will get 420mL of 5% solution.

So, to recap, we started with 250mL of 8.4% solution, which means that 91.6% (229mL) of that solution was pure water and 21mL was pure bicarbonate. We increased the water content to 229+170=399mL, without increasing the bicarbonate content (stays at 21mL). So the volume of solution is 399+21=420mL, of which (21/420)×100=5% is bicarbonate.

If we pour out 250mL of this new solution it will contain 5% (12.5mL) of pure bicarbonate, so we will have lost the required amount of bicarbonate (from 21mL to 12.5mL).