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+?

Answer 1

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

then divide the moles by total volume to get the concentration.

in the end, you will find out that [M2+]= 0.04 M

hope this helps

Answer 2

Answer:

0.02 M

Explanation:

10.0 ml of a 0.1 M of M2+ means that, at the beginning, there were present:

10 ml * 0.1 mol/1000 ml = 0.001 mol of M2+

Analogously, there were 0.001 mol of L

After the reaction occurred the concentration of L is 0.01M. The volume of the solution is now 10 ml + 10 ml = 20 ml, this means that there is present:

20 ml * 0.01 mol/1000 ml = 0.0002 mol of L

So, 0.001 - 0.0002 = 0.0008 mol of L reacted

From the balanced equation:

M2+ (aq) + 2 L (aq)<-> ML2 2+ (aq)

we know that 2 moles of L reacts to form 1 mole of ML2 2+, then 0.0008 mol of L produce:

0.0008 mol of L / 2 moles of L = x  mole of ML2 2+/ 1 mole of ML2 2+

x = (0.0008/2)*1

x = 0.0004 moles of ML2 2+

The concentration of ML2 2+ is the number of moles divided by the volume of the solution, that is:

concentration: 0.0004 moles of ML2 2+/ 0.02 L = 0.02 M

(20 ml is equivalent to 0.02 litre)