[alexnkc]

Hi all,

I am new here and I hope I am posting the question on the right place (please suggest to me if I am posting on the wrong forum. Thanks)

Question 01.

Existing solution level: 300 L
Existing concentration: 250 mL/L (let say 250 mL of solution A in 750 mL of Water)
*unknown detail of solution A

How much of the solution I need to add to increase the existing concentration of 250 mL/L to 300 mL/L

Thanks in advance for the help.

Best regards,
KC

[Hunter2]

You didnt tell the concentration of the solution which you want to add. Your existing bath contains 250 ml/l and this should be increased to 300 ml/l. You can calculate how much is the difference. You know the volume so you know how much it has to be increased, but without a concentration number of your additive you cannot solve it or is the additive 100%?

[alexnkc]

Hi Hunter2,

In our chemical system, we are controlling the concentration base on vol. per vol. (Volume of solution A per volume of total solution). This is because we don't have the solution A concentration.

With referring to Question 1.
300 mL/L - 250 mL/L = 50 mL/L (Each little lack of 50 mL of solution A)

Bath level = 300 L. Therefore, I need to add 15 L of solution A (300 L x 0.05 L).
But I guess above method will not achieve 300 mL/L.

Please help on above matter.

Thanks

Best regards,
KC Ng

[Arkcon]

I'd like to welcome you, alexnkc: to the Chemical Forums.  According to the Forum Rules{click}, we want to see your attempt.   You've posted essentially the same question twice now, but you seem not to be able to see that without some work on your part, this problem won't be solvable.  I suspect we're facing something of a language barrier, but with proper chemical terminology, this particular problem should diminish.

You know the stock concentration, at least in the second question, you know initial and final volumes in your tank.  If you know how much stick solution you need, and how much additional volume you need, you should be able to figure it out.  Consider: what if you had simply an empty 50 l tank to fill?  Isn't that the same problem as increasing the fill of a tank?

[Dan]

Also,

Read this: http://en.wikipedia.org/wiki/Dilution_%28equation%29

[Hunter2]

Hi Hunter2,

In our chemical system, we are controlling the concentration base on vol. per vol. (Volume of solution A per volume of total solution). This is because we don't have the solution A concentration.

With referring to Question 1.
300 mL/L - 250 mL/L = 50 mL/L (Each little lack of 50 mL of solution A)

Bath level = 300 L. Therefore, I need to add 15 L of solution A (300 L x 0.05 L).
But I guess above method will not achieve 300 mL/L.

Please help on above matter.

Thanks

Best regards,
KC Ng

This will guide to to much volume and your content is 90 l/315 l = 285 ml/l

No the Equation has to be V₃₀₀*300 m/l = (V₃₀₀ - V(dumb))*250 ml/l + V(add)(100%)

Vadd = Vdumb

V₃₀₀*300 m/l = (V₃₀₀ - V(dumb))*250 ml/l + V(dumb)

90 l = 75 l -250 ml/l V(dumb) + V(dumb)

15 l = 0,75 l/l * (Vdumb)

V dumb = 20 l

So you have to dumb 20 l of the solution you remain 280 l with the concentration of 250 m/l means 70 l concentrate containing. Now you have to add 20 l concentrate and your Volume is back to 300. You have now 70 l +20 l = 90 l concentrate what means you have 90l/300l = 300 ml/l

[alexnkc]

Hi Hunter2,

Thanks for your guide. Very clear explanation to me.
Thanks again

Best regards,
KC