The equivalence point of a titration is reached when 35.50 mL of 0.40 M HBr is added to a 25.00 mL sample of LiOH. The original [LiOH] is...
a. 0.014 M
b. 0.57 M
c. 0.024 M
d. 0.28 M
M acid x L acid = M base x L base
Substitute and solve for the unknown.
To find the original concentration of LiOH, we can use the concept of stoichiometry in a reaction. The balanced chemical equation for the reaction between HBr and LiOH is:
HBr + LiOH → LiBr + H2O
We can see that the mole ratio between HBr and LiOH is 1:1. Since we know the volume and molarity of HBr added, we can use this information to calculate the moles of HBr added.
Step 1: Calculate the moles of HBr added.
Moles of HBr = volume (in liters) × molarity
= 0.0355 L × 0.40 mol/L
= 0.0142 mol
Step 2: Since the mole ratio between HBr and LiOH is 1:1, the moles of HBr added is equal to the moles of LiOH present in the original solution.
Step 3: Calculate the original concentration of LiOH.
Original concentration of LiOH = moles of LiOH / volume (in liters)
= 0.0142 mol / 0.025 L
= 0.568 M
Therefore, the original [LiOH] is approximately 0.57 M. Option b. 0.57 M is the correct answer.
To find the original concentration of LiOH, we first need to identify the balanced chemical equation for the reaction between HBr and LiOH:
HBr + LiOH → LiBr + H₂O
From the equation, we see that the stoichiometric ratio between HBr and LiOH is 1:1. This means that it takes the same number of moles of HBr and LiOH to react completely.
Given that the equivalence point of the titration is reached when 35.50 mL of 0.40 M HBr is added to a 25.00 mL sample of LiOH, we can calculate the number of moles of HBr used:
moles HBr = volume of HBr (L) × concentration of HBr (mol/L)
= 35.50 mL × (1 L/1000 mL) × 0.40 mol/L
= 0.0142 mol
Since the stoichiometric ratio between HBr and LiOH is 1:1, the number of moles of LiOH used is also 0.0142 mol.
Now, we can calculate the original concentration of LiOH using the original volume of LiOH:
original concentration of LiOH (mol/L) = moles of LiOH / original volume of LiOH (L)
= 0.0142 mol / (25.00 mL × (1 L/1000 mL))
= 0.568 mol/L
Therefore, the original concentration of LiOH is approximately 0.57 M.
The correct answer is b. 0.57 M.