A student placed 10 g of Mg(OH)2 into a flask and added 500ml of water. The student took 100 ml of the Mg(OH)2 solution and reacted with HCl(aq). If 13.75 ml of 0.010M HCL reacted with the OH(aq), was was the concentration of hydroxide ion in the 100 ml solution?

2HCl + Mg(OH)2 ==> MgCl2 + 2H2O

moles HCl = M x L
moles Mg(OH)2 = moles HCl x (1/2)
moles OH^- = 2 x moles Mg(OH)2.

To find the concentration of hydroxide ion (OH-) in the 100 ml solution, we need to use the stoichiometry of the reaction between Mg(OH)2 and HCl.

From the balanced equation of the reaction:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O

We can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl to produce 2 moles of water. This means that the ratio of Mg(OH)2 to HCl is 1:2.

Given that 13.75 ml of 0.010 M HCl reacted with the OH(aq), we can calculate the moles of HCl used:
moles of HCl = volume of HCl (in L) x concentration of HCl

Converting the volume from ml to L:
13.75 ml = 13.75 / 1000 L = 0.01375 L

Calculating the moles of HCl:
moles of HCl = 0.01375 L x 0.010 M = 0.0001375 moles

Since the ratio of Mg(OH)2 to HCl is 1:2, the moles of OH- that reacted will also be 0.0001375 moles.

Now, let's consider that the 0.0001375 moles of OH- reacted in 100 ml of solution.

To find the concentration of OH-, we need to divide the moles of OH- by the volume of the solution in liters:

concentration of OH- = moles of OH- / volume of solution (in L)
= 0.0001375 moles / 0.100 L
= 0.001375 M

Therefore, the concentration of the hydroxide ion (OH-) in the 100 ml solution is 0.001375 M.