The solution should be diluted to 2 L to reduce the concentration of the hydrogen ion, [H+] to one-half of that in the original solution.
Let the original molarity of the solution be 1 M
Thus,
The molarity of the diluted solution = 1 × ½ = 0.5 M
Volume of stock solution (V₁) = 1 L
Molarity of stock solution (M₁) = 1 M
Molarity of diluted solution (M₂) = 0.5 M
1 × 1 = 0.5 × V₂
1 = 0.5 × V₂
Divide both side by 0.5
V₂ = 1 / 0.5
Therefore, the solution should be diluted to 2 L to reduce the concentration of the hydrogen ion, [H+] to one-half of that in the original solution.
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