In a particular titration experiment

Chemistry
Tutor: None Selected Time limit: 1 Day

In a particular titration experiment a 25.0 ml sample of an unknown diprotic acid required 15.0 ml of 0.200 M NaOH for the end point to be reached.  What is the concentration of the acid?

0.167 M

0.240 M

0.0600 M

0.0159 M

Sep 1st, 2015

Thank you for the opportunity to help you with your question!

The equation you are looking for  is v1m1 =  v2m2 

Therefore,   25 ml  *  M1 =  15 ml * .200 ml 


Step one   

Multiply the right side ( 15ml * .200ml) 

25 ml * M1  =    3 ml 

Step two 

Divide each side by 25ml  

120% 

.200 x 120% = .240 



Answer : 0.240 M

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 1st, 2015

I need to talk to you about this question. Are you sure you wrote it down correctly? 







Sep 1st, 2015

15ml * .200 =  3 ml

3ml / 25 =  .120 M 

That is none of the answer choices provided, I misread the problem earlier. 

Sep 1st, 2015

So the answer is .120 or .240?  The .240 is an option on my multiple choice homework but not the .120.

Sep 1st, 2015

After submitting the assignment it turns out the answer is .0600 M so I am still confused about how to solve this problem.

Sep 1st, 2015

I understand. I have contacted my professor to help elborate;however, after discussing the question with my peers and using the equation required (vm=m1m1), we picked the same answer.

Sep 1st, 2015

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