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