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Need help with this molarity problem that involves a neutralization reaction.

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mummer43:
I have no idea how to do this problem!

What volume (L) of 0.0250 M HNO3 is required to neutralize a solution prepared by disolving 17.5g of NaOH in 350 mL of water?

HNO3 (aq) + NaOH  -----> H2O (l) + NaNO3

Any help would be greatly appreciated!!

mike:
How many moles of NaOH are there in the solution?

metalriffzz:
Well to get you started out this is a problem where you're going to have to neutralize the moles of NaOH you have with an equal number of moles of HNO3. Your molar weight for NaOH is roughly 40 g/mol and rember that molarity = moles per liter, meaning that for .0250 M HNO3 there's .0250 moles of acid per liter of solution you add.

17.5/40 = ?

? = ??(.0250)

Good luck.

Borek:
Quite a lot - almost two buckets!

Are you sure you have to use such diluted acid?

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