If it takes 50 mL of 0.05M HCL to neutralize 345 mL of Mg(OH)2 solution, what is the concentration of the Mg(OH)2 solution?

First, write the unbalanced equation:

`Mg (OH)_2 + H Cl = Mg Cl_2 + H_2 O.`

It is simple to balance it:

`Mg (OH)_2 + 2H Cl = Mg Cl_2 + 2H_2 O`

(give the coefficient `2` to `H Cl` to balance `2 Cl` at the right side, then give the coefficient `2` to `H_2 O` to balance `O,` and we are done).

Thus one mole of `Mg (OH)_2` requires two moles of `H Cl` to be neutralized.

Next, there are `0.05 (mol)/L * 50*10^(-3) L = 2.5*10^-3 mol`  of  `H Cl`  in  `H Cl` solution, therefore there are twice less moles of `Mg (OH)_2,` `1.25*10^-3 mol.` This gives the molarity (molar concentration) of `Mg (OH)_2` to be equal to  `(1.25*10^-3 mol)/(345 mL) = (1.25 mol)/(345 L) approx 0.0036 M.`  This is the answer.