[simon_v]

Ok - I'm stuck  

This is the question:

"50ml of 0.2M NaOH was mixed with 50ml of 0.30M MgCl₂. A ppt of Mg(OH)₂ formed. Calculate the conc. of:

a. Na⁺
b. Mg⁺
c. Cl^- "

This is what I've come up with so far:

For the reaction:

2NaOH _(aq)+MgCl_2(aq) > 2NaCl_(aq)+ Mg(OH)_2(s)

The net ionic eq'n is:

2Na⁺_(aq) + 2OH^-_(aq) + 2Mg²⁺_(aq) +2CL^-_(aq) > 2Na⁺_(aq) + 2Cl^-_(aq) + Mg(OH)_2(s)

um....

so that means that all of the Na and Cl ions will be present in the products as they were in the reactants and the Mg and OH ions will ppt out right?

So given that C₁V₁=C₂V₂...

for Na ions:

conc. Na⁺: 0.05L x 0.2M/0.1L (the new volume when 50ml of one is added to 50ml of the other) = 0.1M Na⁺

and for Cl ions:

conc. Cl^-: 2(0.05L x 0.3M/0.1M) = 0.30M Cl^-

now this is where I'm stuck...

I thought the amount of Mg ions would be calculated the same way giving an answer of 0.15M Mg²⁺ (because of 1:2 of Mg:Cl ions in Mg Cl₂)but the answer in the book says 0.1M - is this correct? If so how did they get it? Do I have to work out the moles of Mg(OH)₂ to work out the conc of Mg ions?

[AWK]

Some Mg(2+) precipitated(0.005 mole), hence it exists in a precipitate, not in solution. The answer in textbook is correct.

[simon_v]

I just figured this out - I was looking at it the wrong way assuming that neither were in excess when in actual fact the MgCl₂ is in excess and not all the Mg will ppt out leaving some in solution.