URGENT Concentration of Vinegar | Titration? [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 3 years ago.








[Improve this question](/posts/127912/edit) 



Example of acid base titration




The moles of acid and base are equal in the end point, so $n\_A=n\_B$ where $A$ and $B$ are acid and base, respectively. The number of moles can be written as $n=C\*V$ where $C$ is the molar concentration and $V$ is the volume. So


$n\_A=n\_B$


$C\_A\*V\_A=C\_B\*V\_B$


$C\_A\*10mL=15.7 mL\*0.05 M$


Threfore


$C\_A= 0.0785M$


Now, this is moles of acetic acid per liter. The mass per mole of acetic acid is $M\_r=60,052 g/mol$. So in grams per liter this is


$0.0785M\*60.052 g/mol= 4.71 g/L$


So now you have 4.71 g in 1000 mL (1L). The only thing left to do is a simple rule of three because %m/V es mass of acetic acid in 100 mL, not 1000 mL. Then


$100\*4.71/1000=0.471$


 2