posted by Sarah Hejnosz .
I'm in 10th grade and we did a titration lab; with every lab we have to answer follow up questions and I'm totally stuck on this one: We did not uses the volume of the water added initially to the Erlenmeyer flask in our calculations. Why?
Ok as far as calculations go, I'm not even sure if I did them right and I just do what my teacher/the book says to do without question so I have no idea why I didn't use the volume of water in the flask for the calculations. Is it just because they were irrelevant? Sorry if this made no sense, it's due tomorrow so any help is greatly appreciated!
It isn't irrelevant and it makes a lot of sense. I have been asked this dozens of time in class and many students wonder how the volume of the water added makes no difference. Two ways to explain it. The most scientific way is this way. Say we titrate 25 mL 1.0 M acid with 25 mL 1.0 M base. What usually happens is that we start with 25.0 mL acid, add some water (not measured of course), add a few drops of indicator solution, then begin titrating with the base. You probably remember that as the end point came closer and closer the person washed down the sides of the flask with MORE water (again not measured) because we don't want drops of acid or base on the walls of the container. Here is why it doesn't matter. Scenario 1.
You have 25.0 x 1.0 = 25 mols of the acid to start. The indicator will turn when exactly 25.0 mL of 1.0 base has been added. It turns when the solution is neutralized. Adding water does nothing to make that neutralization happen. So the indicator is waiting until mols acid = mols base at which point it will change color (at least it will if the equivalence point and the end point are the same.) Got it. The end point is when mols acid = mols base and adding water doesn't change that.
Scenario 2 and the one that ALWAYS works with students but it's a little less scierntific.
OK. We've added 25.0 mL of 1.0 M acid to the flask and we add some water before the titration begins. The student asks, "But won't that dilute the acid?" And my answer always is, "Yes, but it will dilute the base (when it is added) exactly the same amount." I hope this helps.
Yes that makes a lot of sense now! Thank you so so much!:)
determine the specific heat capacity of a sample of zinc that weighs 187.2 grams if its temperature is changed from 22.0C to 35.0C when it absorbs 949 J of heat.