Final answer:
The minimum number of half-lives required to reduce the concentration of a 1 M solution to less than 0.05 M in a first-order reaction is 5.
Step-by-step explanation:
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t₁/2 = 0.693/k. The reaction in question is first-order, and we need to find the minimum number of half-lives required to reduce the concentration from 1 M to less than 0.05 M.
Let's use the equation for the half-life of a first-order reaction to solve this: t₁/2 = 0.693/k. We know the concentration after 10 half-lives is 0.053 M, which is greater than 0.05 M, so we need to find how many half-lives it takes to reach less than 0.05 M. Let's set up an equation:
1 M * (1/2)^n = 0.05 M
Solving for n:
(1/2)^n = 0.05
Taking the logarithm of both sides:
n * log(1/2) = log(0.05)
Using the properties of logarithms:
n = log(0.05) / log(1/2)
Calculating:
n ≈ 4.321
Since we can't have a fraction of a half-life, the minimum number of half-lives required to reduce the concentration to less than 0.05 M is 5.