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### Topic: Solution of NaOH ???  (Read 3568 times)

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#### ssstrong

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##### Solution of NaOH ???
« on: June 20, 2008, 12:19:34 PM »
Hi

A Stock solution of 50% NaOH is available . How would 2L of 10% solution be prepared ? Atomic weight of (Na=23,O=16,H=1)

Some people answed me but without using the atomic weight, so what is the advange of them ??

Best reagrds

Saman SarKo

#### Yggdrasil

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##### Re: Solution of NaOH ???
« Reply #1 on: June 20, 2008, 12:47:01 PM »
This is a dilution problem; taking a solution of one concentration and making a solution of a higher concentration.  Given that you know the concentration of the original solution, you don't need to know the molecular weight to solve the problem.

There are two approaches you can take to calculate the volume of 50% NaOH needed for your 10% solution:

1) Calculate a dilution factor.  10%/50% = 1/5, so you are performing a 5x dilution.  This means the volume of 50% NaOH that you use for making your 10% solution is 2L * (1/5) = 0.4L.

2)  Use the equation cinitialvinitial = cfinalvfinal, where c represents the concentrations and v represents the volumes.  This approach is more general and is helpful when the dilution factor is not an easy number to work with (e.g. making a 12.25% solution of NaOH), but mathematically, the two approaches are equivalent.

#### Borek

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##### Re: Solution of NaOH ???
« Reply #2 on: June 20, 2008, 12:54:59 PM »
Ygg: in both cases you are neglecting density changes. They'll be huge in the case of these solutions.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### ssstrong

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##### Re: Solution of NaOH ???
« Reply #3 on: June 20, 2008, 01:07:59 PM »
Thnx dude ...

this question i found it in of the Proff. Lecture ...

May There is another request in the Question which required using the molecular weight ??

Thnx AgAin ..
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