Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: noiseordinance on May 06, 2009, 11:33:48 PM
-
Hi there,
I'm trying to figure out how to determine molarity through titration. So we started with about 250mL of 0.1 M NaOH solution. Then we needed to determine how much KHP would be required to neutralize only 25mL of the 0.1 M NaOH solution, which came out to about 0.51g. Then we used 50mL of NaOH solution and dripped it through a drip counter into a solution of KHP (which also had a pH electrode). It turns out that during the several runs, we had an average of about 24.5mL before the solution became neutral, which tells me (off the top of my head) that the actual molarity of the supposed 0.1 M NaOH is actually less... What I can't seem to figure out is how to tell what the actual molarity is. I also need to provide a standard deviation, which I know how to do... it's just getting the average molarity that's boggling the bajesus outa me. Can anyone provide some help per change? This is probably a pretty basic routine for people who aren't knuckleheads at this stuff like myself.
Thanks!
-
Edit: I think I just figured it out, and if so, it's way easier than I thought. If 25mL was supposed to neutralize the solution, but instead 24.5mL neutralized it, can I just divide 24.5mL / 25mL to find the actual molarity? If 24.5mL / 25.0mL = 0.98, could I then say the molarity is 0.098 M instead of 0.1 M?
-
http://www.titrations.info/titration-calculation
http://www.titrations.info/acid-base-titration-solution-standardization
-
It's very important not to make assumptions no matter how safe they may feel. The data you have should always guide your approach, not expectations. The assumption that 25 mL was supposed to neutralize the solution is the beginning of a dangerous habit that is very easy to form in the structured environment of a classroom-lab. I have a feeling that there is a much deeper lesson to be learned from this lab which relates to the fundamental properties of NaOH and how those properties affect the preparation of an NaOH solution.
-
Edit: I think I just figured it out, and if so, it's way easier than I thought. If 25mL was supposed to neutralize the solution, but instead 24.5mL neutralized it, can I just divide 24.5mL / 25mL to find the actual molarity? If 24.5mL / 25.0mL = 0.98, could I then say the molarity is 0.098 M instead of 0.1 M?
The other way around, perhaps, as for a given mass of analyte, concentration is inversely proportional to volume. 25 ml / 24.5 ml. Assuming of course that other things hold, too.