Chemistry Titration Lab HELP!
posted by Angel on .
Hi everyone! I just got a new lab that I MUST complete today, and it's about titration I have absolutely no clue how to do it, it just doesn't make sense to me! I hope that someone can help explain to me how to do it so I can finish ASAP!!
Thanks a lot,
Angel
LabTitrations
Questions:
Use 40 ml of KOH for your calculations.
1. Calculate the concentration of the HCl acid sample you titrated in this lab for two of the trials you ran that did not have an overshot endpoint. You will be reporting two answers. You can do your calculations on a piece of scratch paper.
Trial 1 concentration:
Trial 2 concentration:
2. Explain how you arrived at your answers.
3. Imagine that you ran a similar titration using HCl as the sample with 3.56 M KOH as the titrant. Calculate the concentration of the HCl if the sample size were 25 mL, and if the volume of the titrant used were 29.6 mL.
4. Calculate the concentration of a 50.0 mL sample of HBr acid, which was titrated with 37.7 mL of 0.57 M NaOH base.
5. Explain the difference between the equivalence point and the endpoint of a titration.
6. Explain why an overshot endpoint is not a good titration.

Go to google and type in titrations lab.
The 6th result "Performing Titration Lab" has lots of information and even a virtual titration tool that you can use to get the hang of it! 
Oh, PERFECT! Thanks you so much! I'll check the 7th one right now!

My mistake, the 7th result.
Chemlab_experiment 
LabTitrations
Questions:
Use 40 ml of KOH for your calculations.
1. Calculate the concentration of the HCl acid sample you titrated in this lab for two of the trials you ran that did not have an overshot endpoint. You will be reporting two answers. You can do your calculations on a piece of scratch paper.
Trial 1 concentration:
Trial 2 concentration:
We can't help on this first one because you don't list any values for the titration. And it's unclear what the 40 mL KOH has to do with it.
2. Explain how you arrived at your answers.
Same response for #2. No data.
3. Imagine that you ran a similar titration using HCl as the sample with 3.56 M KOH as the titrant. Calculate the concentration of the HCl if the sample size were 25 mL, and if the volume of the titrant used were 29.6 mL.
Here you have a titration using 25 mL HCl and you titrate it with 29.6 mL of 3.56 M KOH. In all of these calculations, just remember a couple of things. One is the definition of molarity. M = mols/L which can be rearranged to mols = M x L or L = mols/M. The second, when mass is involved is mols = grams/molar mass. For this problem, how many mols KOH were used in the titration? That is mol=M x L = 3.56 x 0.0296 L = 0.1054 mol KOH. Since KOH and HCl react 1:1 in a balanced equation (you may want to write the equation), that means we have 0.1054 mol HCl at the equivalence point. So what is the molarity of the HCl? M = mols/L = 0.1054 mol/0.025 L = 4.215 M which I would round to 4.22 M for HCl.
4. Calculate the concentration of a 50.0 mL sample of HBr acid, which was titrated with 37.7 mL of 0.57 M NaOH base.
Done exactly the same way as problem 3.
5. Explain the difference between the equivalence point and the endpoint of a titration.
The equivalence point is the exact point at which the acid/base system is neutralized. The end point is where the indicator shows that the equivalence point has been reached. The end point and the equivalence point will be almost the same IF the indicator has been chosen properly.
6. Explain why an overshot endpoint is not a good titration.
Isn't this an obvious answer 
Oh, thanks SO much DrBob222! I'm going to go step by step the way you dictated it! I'm going to try and work it out right now!

Okay, for number 1 and number 2, they were both for an experiment that was online, sorry for the confusion I think I'll be fine answering it! If anything, I'll ask about it with specific detail thank you guys!

You're welcome.

i have a titration lab with sodium hydroxide with an unknown vinegar
all i know is we have 0.3 M sodium hydroxide solution (250 mL)and a 50ml burette that when titration was done the avg. of 3 trials was 27.7 mL
im not sure what id write for the hypothesis of this lab or what a balanced chemical equation would be
help please!! 
Okay, finished some Chem quizes, and now here's number 3, with DrBob222's directions (hope I did this right:
Titration using 50mL of HBr acid with 37.7mL of 0.57 M NaOH base.
Home many M NaOH base in titration
(mol= M * L)
0.57 * 0.0377 L= 0.21489mol NaOH base.
Molarity of HBr acid:
(M=Mols/L)
M HBr= 0.021489/0.050L= 0.42978 M
Thanks a lot everyone!
Angel 
what's suggest an application for acid/base titration

Keep it simple please

hey ur website is very confusing