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Offline nicholasbellono

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titration
« on: October 06, 2007, 02:05:17 PM »
I really appreciate the help on here in my previous questions. I am new to chemistry and the direction I have recieved has been very beneifical to me moving forward with my lessons.
I have one more question from my lab notes...

a 25.00 mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125 mL Erlenmeyer flask and 2 drops of phenolphthalein are added. The above sodium hrydroxide soltuion (the titrant) is used to titrate the nitric acid acid solution (the ayalyte). If 12.75 mL of the titrant is dispensed from a buret in causing a color change of the phenolphthalein, what is the molar concentration of the nitric acid solution?

I calculated the concentration of the titrant for the above question (NaOH) is .16M from this question:
4g NaOH is dissolved in 5 mL of water. a 4 mL aliquot of this solution is diluted to 500 mL of solution. What is the approximate molar concentration of NaOH in the diluted solution?

I'm not positive on the .16M

How do i set up this calculation???

I thought for a molar concentration of acid it was mol acid/ volume acid. but that doesn't seem to fit.

Thanks!

Offline Yggdrasil

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Re: titration
« Reply #1 on: October 06, 2007, 03:25:14 PM »
I also get 0.16M as the concentration of the NaOH solution.

Here are my suggestions on how to start.  First, always start with a balanced chemical reaction.  Next, find out the number of moles of sodium hydroxide added to the solution.  Then, from the balaned chemical reaction, you can figure out the number of moles of nitric acid that were in your 25mL aliquot.

Offline nicholasbellono

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Re: titration
« Reply #2 on: October 06, 2007, 03:37:26 PM »
i did mL X mol = mL X mol
25mL X mol HNO3 = 12.75 X .16M NaOH
molar concentration= .08160

what about when it says dissolved with instead of disolved in?

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