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How many milliliters of 0.0991 M LiOH are required to titrate 25.0 mL of HCl to the equivalence point

User Doniel
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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.
User Hainq
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