31.0mL sample of LiOH, concentration unknown, is titrated with .52 M HCl. It takes 42.9 ml of HCl to reach the endpoint. What is the concentration of the base?
mL acid x M acid = mL base x M base.
0.71 M
I don't think you rounded properly. My calculator tells me 0.71961 which I would round to 0.72 to 2 s.f. and to 0.720 to 3 s.f.
ok thanks!
To find the concentration of the LiOH base, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction between LiOH and HCl:
LiOH + HCl -> LiCl + H2O
From the balanced equation, we can see that the mole ratio between LiOH and HCl is 1:1. This means that 1 mole of LiOH reacts with 1 mole of HCl.
Given that it took 42.9 mL of 0.52 M HCl to reach the endpoint, we can determine the number of moles of HCl used in the titration.
Moles of HCl = concentration (M) x volume (L)
= 0.52 M x (42.9 mL / 1000 mL/L)
≈ 0.022 mol HCl
Since the mole ratio between LiOH and HCl is 1:1, the number of moles of LiOH is also 0.022 mol.
Now, let's calculate the concentration of the LiOH solution.
Concentration of LiOH (M) = moles / volume (L)
= 0.022 mol / (31.0 mL / 1000 mL/L)
≈ 0.710 M
Therefore, the concentration of the LiOH base is approximately 0.710 M.