[Ishikaya]

During a fabrication of a batch of NaOH which was supposed to be 36 600g of NaOH added to 238Kg of water. Instead 36 600g were added to 260Kg of water. Since the batch makers work in mass the solution ended up being 296.6Kg instead of 274.6 so they dumped out 22Kg of the solution. So here is the question how much NaOH do I add to the batch we have to obtaine the same amount of NaOH we are supposed to have?

[Arkcon]

Do a little bit of math. 36 000 g in 296.6Kg total mass is what molality (abbreviated m) solution?And the molaity for the correct mix of 36 000 g in  274.6Kg is what?  Once you know that setup a proportion, and add water.

Do you have a QC test to verify if its made properly in any case?  If so, you can then check your adjusted batch.

Sorry to answer a question with a question, but that's what we do here.  High School student or production engineer, everyone follows the Forum Rules{click}.

[Ishikaya]

see that is were it gets a little complicated since the 296.6kg includes the mass of NaOH shouldnt we be doing 36 0000g/260Kg giving us 3.46N. The concentration I want is 3.84N I know this ... im actually looking to verify my calculations

[Arkcon]

Don't use molarity or normality(N) in this case.  Use molaity, because you're using a certain mass of solid in a certain mass of total solution.  Since you were using that unit anyway, by virtue of using final mass as endpoint check, its better for you if you do it using molality.

[Ishikaya]

Alright here is my calculation please verify:

Solution should be = 36 600g + 238Kg giving us a concentration of 3.33 mol/Kg

Original solution we had = 36 600g + 260Kg giving us a concentraion of 3.08 mol/Kg

* at this point 22Kg of solution was removed thus we are left qith 274.6kG of solution at a concentration of 3.08 mol/Kg

* thus I have to add 0.25 mol/Kg of solution which is 2 746g

Is this correct

and thank you

[AWK]

It is not correct.

* at this point 22Kg of solution was removed thus we are left qith 274.6kG of solution at a concentration of 3.08 mol/Kg

Water was removed, not solution.

This is a simplest problem for using mass fraction. You should set one equation with one variable and solve it. This is an equivalent to percentage calculations.

For molal calculation you should set the analogous equation with mass od water in denominator.

This is not clear which one type of calculation is needed from your question

So here is the question how much NaOH do I add to the batch we have to obtaine the same amount of NaOH we are supposed to have?

[Arkcon]

Also, what formula weight are you using for NaOH?  I'm not getting the moles you report for those masses.