[Misslizajane]

   Hey you all. I have tried to set up my practice problem several times and I cannot get correct answer the book has which is 60.0 g/mol.   It's a practice question so your not doing any homework for me.  I have just been working on this problem all afternoon and I am truely stumped and I'm curious to see what I'm failing to do to solve it. lol. Thanks for your time if you decide to take this question on.   

Answer in back of book is 60.0 g/mol.    but I cannot get it

The Question is:   A 0.125-g sample of a monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH.  The endpoint is reached after adding 20.77 mL of base.  What is the molar mass of the unknown acid? 

[Astrokel]

Show your working and we can tell you where you went wrong.

[Misslizajane]

I went to work on it again and this is what I got.


I take my 0.1003 M NaOH and multiply it by 20.77 mL (which I will convert to L)

(0.1003 M NaOH)(.02077 L)= .002083 

Then I take the .125 g sample of unknown acid and divide that by .002083

.125/.002083  = 60.0 g/mol of uknown acid!!!!

I did it!!! After pulling out all my hair!!! 

[Misslizajane]

Ok, So I figured out the mass of the unknown monprotic acid in above posts.  Now I am tryin to figure out if unknown mass of diprotic acid.  I am just wondering does diprotic acid make any diffrence on how I solve my problem or does it mean there is a 2 to 2 mole ratio of the acid and base???    Do I go about solving the problem just as I did in my first question.  lol. 

[Astrokel]

I did it!!! After pulling out all my hair!!! 

well done

I am just wondering does diprotic acid make any diffrence on how I solve my problem or does it mean there is a 2 to 2 mole ratio of the acid and base???

Nope the ratio is not 2:2. When in doubt write a balanced equation such as,

H₂A + BOH ---->

complete and balance it and you will see the ratio!

[Misslizajane]

 HEy there. Yr last comment isn't makin sense to me. sorry.

The Question:  A 0.105-g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1288 M NaOH.  The endpoint is reached after adding 15.2 mL of base.  What is the molar mass of the unknown acid? 

Given: .105 g of diprotic acid
Given: .1288 M NaOH
Given: 15.2 mL= .0152 L
Find: Molar mass of unknown acid

So I took:   (.1288 M NaOH)(.0152 L)  = .00195776 which I rounded to 3 significant figures .00196 

Then I took my  .105g of diprotic acid and divided it by the .00196
.105/.00196= 53.57 g/mol.



I have a feeling that is wrong.  I have a feeling like I'm supposed to multiply something by 2.  What am I missing?

     

[Astrokel]

Hey, you are wrong because you assume the ratio of acid reacting with base is 1:1. However in diprotic acid it is not. My previously comment was trying to help you to balance the reaction between diprotic acid and NaOH so you could see the stoichiometry ratio. Perhaps you would want to try balancing again?

H₂X + NaOH ---->

[Misslizajane]

thank you very much. I think it's just been lack of sleep...ha!ha!  maybe one day I will be  a pro with practice and practice. Again thank you so much for your help and encouragement! You are too kind!

Miss Liza Jane