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+?

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)