Final answer:
To neutralize 8L of 2M H₂SO4, you will need 2L of 4M Mg(OH)₂.
Step-by-step explanation:
To find the amount of 4M Mg(OH)₂ needed to neutralize 8L of 2M H₂SO4, you can use the formula CV = C₂V₂. Here, C represents the concentration and V represents the volume. First, determine the moles of H₂SO4 using the formula moles = concentration × volume. So, moles of H₂SO4 = 2M × 8L = 16 moles. Since the ratio between Mg(OH)₂ and H₂SO4 is 2:1, you will need half the amount of moles of Mg(OH)₂. So, moles of Mg(OH)₂ = 16/2 = 8 moles. Now, use the formula V₂ = moles ÷ concentration to find the volume of 4M Mg(OH)₂. Here, moles = 8 and concentration = 4M. So, V₂ = 8 moles ÷ 4M = 2L.
Therefore, you will need 2L of 4M Mg(OH)₂ to neutralize 8L of 2M H₂SO4.
Learn more about Neutralization of acids and bases