Question

What volume (in mL) of 0.3250 M HCl is required to neutralize 75.00 mL of 0.6000 M LiOH?

a) 32.50 mL
b) 75.00 mL
c) 150.00 mL
d) 25.00 mL

Answer 1

Neutralize 75.00 mL of 0.6000 M LiOH with 0.3250 M HCl, we can use stoichiometry. The volume of 0.3250 M HCl required is approximately 0.138 mL.

Step-by-step explanation:

To determine the volume of 0.3250 M HCl required to neutralize 75.00 mL of 0.6000 M LiOH, we can use the concept of stoichiometry. The balanced chemical equation for the reaction is:

HCl + LiOH → LiCl + H2O

From the equation, we can see that the ratio of HCl to LiOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of LiOH. We can calculate the moles of LiOH using the formula:

moles = concentration * volume (in liters)

moles of LiOH = 0.6000 M * 75.00 mL / 1000 mL/L = 0.0450 moles

Since the ratio is 1:1, the moles of HCl required will also be 0.0450 moles. We can use the formula to calculate the volume of HCl:

volume of HCl = moles of HCl / concentration of HCl

volume of HCl = 0.0450 moles / 0.3250 M = 0.138 mL

Therefore, the volume of 0.3250 M HCl required to neutralize 75.00 mL of 0.6000 M LiOH is approximately 0.138 mL.