The balanced chemical equation for the reaction between LiOH and HCl is:
LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l)
From the equation, we can see that the stoichiometry of the reaction is 1:1, which means that 1 mole of LiOH reacts with 1 mole of HCl.
To determine the volume of LiOH required to titrate 25.0 mL of 0.0991 M HCl, we need to use the equation:
Molarity × Volume = moles
First, we need to calculate the number of moles of HCl in 25.0 mL of 0.0991 M HCl:
Moles of HCl = Molarity × Volume = 0.0991 mol/L × 0.0250 L = 0.00248 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of LiOH required to react with the HCl is also 0.00248 mol.
Now we can use the same equation to calculate the volume of 0.0991 M LiOH required to provide 0.00248 mol of LiOH:
Molarity × Volume = moles
0.0991 mol/L × Volume = 0.00248 mol
Volume = 0.00248 mol ÷ 0.0991 mol/L = 0.0250 L = 25.0 mL
Therefore, 25.0 mL of 0.0991 M LiOH are required to titrate 25.0 mL of HCl to the equivalence point.