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.
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Hi Ma'am :)
But sorry i learnt about equlibrium last year ,i completely forgot
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I Hope someone will help you out soon =)
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.
That is a good start. I will redo the calculations.
That's good that you have a start :) !
Thanks for trying Hershey. The important point that I got from you is that I have to redo the calculations with the new volumes.
If I did it correctly, the answer should've been [M2+] = 0.04 M.
That is what my nephew got, but the teacher says otherwise. Thanks Rogue!