- #1
Radja24
- 7
- 0
Hi, I'm taking AP Chem this year, and we just had this big assignment. I got through most of it, but here's some I need a walkthrough with.
A titration is done using .1302M NaOH to determine the molecular weight of an acid. The acid contains one acidic hydrogen per molecule. If 1.863 grams of the acid requires 70.11mL of the NaOH solution, what is the molecular weight of the acid?
Here's the second one:
Fe (II) + Ce (IV) -----> Fe (III) + Ce (III)
What is the concentration of Fe(II) if it requires 21.35mL of .3136 M Ce (IV) to titrate a 10.00mL aliquot to the endpoint?
Attempt:
1st one, I tried to get moles of NaOH and then multiply it by 110 to get 1 mole, then multiply 1.863 by 110...but that makes no sense.
For the second, I tried using MfVf formula for titration, and before that I tried to get the moles of Ce (Iv) once again, and then find the limiting agent and plug and go from there, but I sincerely am stuck.
These were two ones I couldn't get really get after some nonsense work. Please help...
Homework Statement
A titration is done using .1302M NaOH to determine the molecular weight of an acid. The acid contains one acidic hydrogen per molecule. If 1.863 grams of the acid requires 70.11mL of the NaOH solution, what is the molecular weight of the acid?
Here's the second one:
Fe (II) + Ce (IV) -----> Fe (III) + Ce (III)
What is the concentration of Fe(II) if it requires 21.35mL of .3136 M Ce (IV) to titrate a 10.00mL aliquot to the endpoint?
Attempt:
1st one, I tried to get moles of NaOH and then multiply it by 110 to get 1 mole, then multiply 1.863 by 110...but that makes no sense.
For the second, I tried using MfVf formula for titration, and before that I tried to get the moles of Ce (Iv) once again, and then find the limiting agent and plug and go from there, but I sincerely am stuck.
These were two ones I couldn't get really get after some nonsense work. Please help...
Last edited: