Jump to content

I need help with NaOH titrations


Recommended Posts

Erm, unless im not reading this correctly, it doesn't make sence. You said that the sodium hydroxide hadn't dissolved completely and then she did the titration with an acid to determine the molarity. But then you say that she leaves it a week, by which time the sodium hydroxide has completely dissolved and then you say she does the titration...so when did she do the titration, because that is important.


If she did the titration when the sodium hydroxide hadn't all dissolved, then yes she will have a molarity lower than she should. The titration only determines the concentration in solution. However, I would expect that upon addition of the acid to the solution, it would make the sodium hydroxide dissovle faster.


If she did the titration the week latter once it had all completely dissolved, then it would make no difference, she would get the right molarity.

Link to comment
Share on other sites

Yeah I am having the same problem in understanding the question. Did you check the molarity straight away when there was still undissolved solid? If that is the case (why on earth did you do that lol) then you will need to perform a new titration to find the new concentration since it has fully dissolved.



Ooops, wait scratch that, I think I may understand. There are two completely different sets of titrations going on here. The first was before the solid was dissolved and it was titrated against an acid of known conc. to find the molarity of your NaOH? Now a week later you are using the NaOH for a second series of titrations but assuming you have NaOH of known conc. and working from there?


If that is the case then unfortunately your solution is now more concentrated than you had been assuming for your experiments. Now I can't really say for certain what you should do without understanding in more detail what your work involves, but don't worry too much as this doesn't necessarily mean you have wasted all the previous experiments. If you titrate your NaOH again with an acid of known conc (so long as you are positive that it was all disolved at the time of beginning your work a week later) and then assume that this was the conc. you used for all the more recent experiments. Incase it isn't immediately obvious simply find the percentage concentration the old solution was of the new solution (boy thats an awful sentence haha) and extrapolate fromvthere. For example say you found the old one was 1moldm-3 and the fully dissolved one is 2moldm-3 you take the value of the titre (in moles) and divide it by 50 and then multiply it by 100 and unless I am being silly that should give you the actual number of moles involved in the reaction. Remember to do this with moles not ml haha.

Edited by farmboy
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.