10.0 ml of a 0.1 M solution of a metal ion solution of a Metal ion M2+ is mixed with a 10.0 ml Solution of a 0.1 M substance L. The following equation is established: M2+(aq)+2L(aq)ML2 2+(aq). At equilibrium the concentration of L is 0.01M. What is the equilibrium concentration of [ML2]2+? Please help me understand what I am doing wrong.

Question

10.0 ml of a 0.1 M solution of a metal ion solution of a Metal ion M2+ is mixed with a 10.0 ml Solution of a 0.1 M substance L.  The following equation is established: M2+(aq)+2L(aq)<->ML2 2+(aq). At equilibrium the concentration of L is 0.01M.  What is the equilibrium concentration of [ML2]2+?  Please help me understand what I am doing wrong.

preetha · Answer

Hi Diva

preetha · Answer

Diya

diyadiya · Answer

Hi Ma'am :)
But sorry i learnt about equlibrium last year ,i completely forgot

diyadiya · Answer

I Hope someone will help you out soon  =)

anonymous · Answer

i know how to solve 10.0 mL of a 0.100 M

it's comes out like this, Hope fully this can give you a little start :


Change everything to moles. Multiply Molarity by Volume to get moles. 

First change L in moles. You can figure out how many moles of each are in equilibrium. 

If L = x moles, M2+ = x moles, ML2 = 3x moles.

PS. To find L moles multiply the final concentration (given above) by the total volume, 200mL.

Divide the moles by the total volume to get the concentration.

preetha · Answer

That is a good start.  I will redo the calculations.

anonymous · Answer

That's good that you have a start :) !

preetha · Answer

Thanks for trying Hershey.  The important point that I got from you is that I have to redo the calculations with the new volumes.

rogue · Answer

If I did it correctly, the answer should've been [M2+] = 0.04 M.

preetha · Answer

That is what my nephew got, but the teacher says otherwise.  Thanks Rogue!

rogue · Answer

Alright, no problem :)