If 37.0 mL of HCl is used to neutralize 25.0 mL of 0.09345M LiOH. What is the concentration of the acid
To find the concentration of the acid, we can use the equation for neutralization reactions:
acid + base → salt + water
In this case, the acid is HCl and the base is LiOH. We are given the volume and concentration of the base (LiOH) and the volume of the acid (HCl) used to neutralize it.
Step 1: Write the balanced chemical equation for the neutralization reaction.
HCl + LiOH → LiCl + H₂O
Step 2: Determine the number of moles of LiOH used.
First, we need to find the number of moles (n) of LiOH used by using its concentration and volume:
n(LiOH) = concentration × volume
n(LiOH) = 0.09345 mol/L × 25.0 mL = 0.09345 mol/L × 0.0250 L = 0.00233625 mol
Step 3: Use stoichiometry to find the number of moles of HCl used.
From the balanced equation, we can see that the stoichiometric coefficient of HCl is 1. This means that 1 mole of HCl reacts with 1 mole of LiOH.
Therefore, the number of moles of HCl used is also 0.00233625 mol.
Step 4: Calculate the concentration of HCl.
We now have the number of moles of HCl and the volume of HCl used:
concentration(HCl) = moles/volume
concentration(HCl) = 0.00233625 mol/37.0 mL = 0.06317961 mol/L
Rounding to the appropriate number of significant figures, the concentration of HCl is approximately 0.0632 mol/L.
Therefore, the concentration of the acid (HCl) is 0.0632 mol/L.