A student took 100 mL of a solution of Mg(OH)2 and reacted it with HCl (aq) to calculate the concentration of hydroxide ions in the solution. If 13.75 mL of 0.010M HCl was used to react with all of the OH- (aq) in the 100-mL solution, what was the concentration of hydroxide ions dissolved in the 100-mL solution?
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
mols HCl = M x L = ?
mols Mg(OH)2 = 1/2 that (use the coefficients in the balanced equation.)
M Mg(OH)2 = mols Mg(OH)2/L Mg(OH)2.
Then M OH^-) = twice M Mg(OH)2 because there are two OH^- ions/Mg(OH)2.
To find the concentration of hydroxide ions in the 100 mL solution, we need to use stoichiometry and the equation balanced with HCl and Mg(OH)2.
Let's analyze the balanced chemical equation:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
From the balanced equation, we can see that there is a 1:2 stoichiometric ratio between Mg(OH)2 and HCl. This means that for every 1 mole of Mg(OH)2, 2 moles of HCl are required to react completely.
We are given that 13.75 mL of 0.010 M HCl was used to react with all the hydroxide ions in the 100 mL solution. To find the moles of HCl used, we can use the following formula:
moles of HCl = volume of HCl (in L) × concentration of HCl (in mol/L)
= 13.75 mL × (0.010 mol/L) / 1000 mL/L
= 0.0001375 mol
Since the stoichiometric ratio between HCl and Mg(OH)2 is 2:1, the moles of Mg(OH)2 in the 100 mL solution is twice the moles of HCl used:
moles of Mg(OH)2 = 2 × 0.0001375 mol
= 0.000275 mol
The concentration of hydroxide ions can now be calculated by dividing the moles of Mg(OH)2 by the volume in liters:
concentration of hydroxide ions (in mol/L) = moles of Mg(OH)2 / volume of solution (in L)
= 0.000275 mol / 0.100 L
= 0.00275 mol/L
Therefore, the concentration of hydroxide ions in the 100-mL solution is 0.00275 M.