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?
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
Multiply the right side ( 15ml * .200ml)
25 ml * M1 = 3 ml
Divide each side by 25ml
120% .200 x 120% = .240
Answer : 0.240 M
I need to talk to you about this question. Are you sure you wrote it down correctly?
15ml * .200 = 3 ml
3ml / 25 = .120 M
So the answer is .120 or .240? The .240 is an option on my multiple choice homework but not the .120.
After submitting the assignment it turns out the answer is .0600 M so I am still confused about how to solve this problem.
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.
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