an acid base titration is conducted in which 20 ml of LiOH is added to 10ml of .200M HBr to reach the equivilance point. what is the molarity of the LiOH solution?
mL x M = mL x M
One unknown.
To determine the molarity of the LiOH solution, we can use the concept of stoichiometry in acid-base titrations.
Given:
Volume of LiOH solution (V1) = 20 mL
Volume of HBr solution (V2) = 10 mL
Molarity of HBr solution (M2) = 0.200 M
At the equivalence point, the moles of acid (HBr) will be equal to the moles of base (LiOH). Therefore, we can use the following equation:
Moles of acid = Moles of base
To calculate the moles of acid, we use the formula:
Moles of acid (HBr) = Volume of acid solution (V2) x Molarity of acid solution (M2)
Moles of acid (HBr) = 10 mL x 0.200 M = 2.00 mmol
Since the moles of acid (HBr) and base (LiOH) are equal, we can calculate the molarity of the LiOH solution using the equation:
Molarity of base solution (M1) = Moles of base / Volume of base solution (V1)
Molarity of base solution (M1) = 2.00 mmol / 20 mL = 0.100 M
Therefore, the molarity of the LiOH solution is 0.100 M.