A student took 100 mL of a solution of Mg(OH)2 and reacted it with H2SO4 (aq) to calculate the concentration of hydroxide ions in the solution. If 13.75 mL of 0.010 M H2SO4 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?
Ca(OH)2 + H2SO4 ==> CaSO4 + 2H2O
mols H2SO4 = M x L = approx 1.5E-4 but that's an estimate and you need a more accurate answer for this and all the calculations that follow.
So mols OH^- must be twice that and (OH^-) = mols OH^-/L. Note L is 0.1 for Ca(OH)2.
To find the concentration of hydroxide ions (OH-) in the 100 mL solution, we can use a simple stoichiometric calculation based on the reaction between Mg(OH)2 and H2SO4:
Mg(OH)2 (aq) + H2SO4 (aq) -> MgSO4 (aq) + 2H2O (l)
From the balanced equation, we can see that 1 molecule of Mg(OH)2 reacts with 2 molecules of H2SO4 to produce 2 molecules of water. This means that the moles of OH- in the solution is equal to half of the moles of H2SO4 used.
First, let's calculate the moles of H2SO4 used:
moles of H2SO4 = concentration of H2SO4 * volume of H2SO4 in liters
= 0.010 mol/L * (13.75 mL / 1000 mL/L)
= 0.0001375 mol
Since we need half of the moles of H2SO4 to obtain the moles of OH-, we can calculate the moles of OH-:
moles of OH- = 0.0001375 mol / 2
= 0.00006875 mol
To find the concentration of OH-, we divide the moles of OH- by the volume of the solution in liters:
concentration of OH- = moles of OH- / volume of solution in liters
= 0.00006875 mol / (100 mL / 1000 mL/L)
= 0.0006875 mol/L
Therefore, the concentration of hydroxide ions in the 100 mL solution is 0.0006875 mol/L or 0.6875 mM.