what volume of 0.4 M HCl is required to neutralize 100.0 mL of 0.3 M LiOH?
For a 1:1 ratio of acid to base, which you have, mL x M = mL x M.
mL(HCl) x 0.4 M = mL(LiOH) x 0.3 M
Solve for mL(HCl).
(ii) If 22.0 mL of 2.5 M HCl was used to neutralize 50.0 mL of a Ba(OH)2 solution, what was the concentration of the base?
75 ml of hcl
0.6
To determine the volume of 0.4 M HCl required to neutralize 100.0 mL of 0.3 M LiOH, we can use the concept of stoichiometry and the equation balanced between HCl and LiOH.
The balanced equation for the neutralization reaction between HCl and LiOH is:
HCl + LiOH → LiCl + H2O
From the balanced equation, we can see that the molar ratio between HCl and LiOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of LiOH.
First, we need to calculate the number of moles of LiOH in the given volume, which is 100.0 mL (0.100 L), by using the formula:
moles = concentration × volume
moles of LiOH = 0.3 M × 0.100 L = 0.030 moles
Since the molar ratio between HCl and LiOH is 1:1, the number of moles of HCl required to neutralize the LiOH is also 0.030 moles.
Now, we can use the formula to find the volume of 0.4 M HCl required to have 0.030 moles of HCl:
volume = moles / concentration
volume of HCl = 0.030 moles / 0.4 M = 0.075 L
Finally, we convert the volume from liters to milliliters:
volume of HCl = 0.075 L × 1000 mL/L = 75.0 mL
Therefore, 75.0 mL of 0.4 M HCl is required to neutralize 100.0 mL of 0.3 M LiOH.