I've been given the Volume of Concentrated NaOh solution (mL) and also the Concentration of Concentrated NaOh solution (M).
I also have the volume of stock solution after dilution (mL).
The question is asking the approximate concentration of stock solution (M)
any help?
5 answers
Your question lacks clarity. Some numbers and how these solutions were prepared would help.
If you know the molarity of the stock solution..you know how many moles you want.
For instance, if you want 0.06M of NaOH of stock solution, 1.400ml of it.
So the moles of NaOH in the stock solution will be .06*1.4 = .084 moles.
You have 12M conc solution.
so you need .084 moles of this.
molarity=moles/volume or
volume=moles/molarity=.084/12=.007 liters, or 7ml.
So 7ml of conc NaOH should be added to sufficient water to make up 1.4liters of solution.
For instance, if you want 0.06M of NaOH of stock solution, 1.400ml of it.
So the moles of NaOH in the stock solution will be .06*1.4 = .084 moles.
You have 12M conc solution.
so you need .084 moles of this.
molarity=moles/volume or
volume=moles/molarity=.084/12=.007 liters, or 7ml.
So 7ml of conc NaOH should be added to sufficient water to make up 1.4liters of solution.
i kinda understand waht you're saying. to be a bit more clear this is from a lab,
and here are the volumes we retrieved:
Volume of Concentrated NaOH soultion (mL) = 9.00 mL
Concentration of Concentrated NaOH soultuon (M) = 6.00 M
Volume of Stock Solution after Dilution (mL) = 490.00 mL + 9.00 mL (from the first part)
and then the question says the give the Approximate concentration of Stock solution (M).
It says to use the formula: amount of solute (moles) / volume of solution (L)
I'm confused with how to get amount of solute with the 6.00 M and 9.00mL volume.
and here are the volumes we retrieved:
Volume of Concentrated NaOH soultion (mL) = 9.00 mL
Concentration of Concentrated NaOH soultuon (M) = 6.00 M
Volume of Stock Solution after Dilution (mL) = 490.00 mL + 9.00 mL (from the first part)
and then the question says the give the Approximate concentration of Stock solution (M).
It says to use the formula: amount of solute (moles) / volume of solution (L)
I'm confused with how to get amount of solute with the 6.00 M and 9.00mL volume.
How many mols NaOH did you start with. M x L = 6.00 M x 0.009 L = ??
When it was diluted you still had ?? mols NaOH. What is the final volume? That is 490.00 + 9.00 mL = 499.00 mL or 0.499 L.
Now M of the diluted solution is mols/L = ??mols/0.499 L. Voila!
When it was diluted you still had ?? mols NaOH. What is the final volume? That is 490.00 + 9.00 mL = 499.00 mL or 0.499 L.
Now M of the diluted solution is mols/L = ??mols/0.499 L. Voila!
OMG, i totally understand that. DrBob222, THANK YOU VEYR MUCH!