A drop of solution volume 0.1 cc contains 6*10^-6 mol of H+ ,if the rate constant of dissappearance of H+ is 10^7 mol/l/sec .how long it take for H+ in drop to dissappear

1)6*10^-3sec

2)6*10^-6sec

3)6*10^-10sec

4)6*10^-9sec

To determine how long it takes for the concentration of H⁺ ions to disappear from the solution, we can use the concept of reaction kinetics. Specifically, we will apply the first-order rate law, which is suitable for reactions where the rate of disappearance of a reactant is proportional to its concentration.

Understanding the Problem

We know the following:

• Volume of the solution = 0.1 cc = 0.1 mL = 0.1 x 10^-3 L
• Amount of H⁺ = 6 x 10^-6 mol
• Rate constant (k) = 10⁷ mol/L/s

Calculating the Initial Concentration

First, we need to find the initial concentration of H⁺ in the solution:

Concentration (C) = Amount of substance (n) / Volume (V)

Substituting the values:

C = (6 x 10^-6 mol) / (0.1 x 10^-3 L) = 6 x 10^-6 / 0.1 x 10^-3 = 6 x 10^-3 mol/L

Applying the First-Order Kinetics Formula

For a first-order reaction, the rate of disappearance of a reactant can be expressed as:

-d[H⁺]/dt = k[H⁺]

Integrating this gives us:

ln([H⁺]₀ / [H⁺]) = kt

Where [H⁺]₀ is the initial concentration and [H⁺] is the concentration at time t.

Finding the Time for Complete Disappearance

To find the time it takes for the H⁺ ions to disappear, we can assume that the concentration drops to a negligible amount. For practical purposes, we can consider it to be close to zero. However, we can also calculate the time for a significant decrease, such as when the concentration drops to 1% of its initial value.

Let’s calculate the time for the concentration to drop to 1%:

1% of [H⁺]₀ = 0.01 x 6 x 10^-3 mol/L = 6 x 10^-5 mol/L

Now, substituting into the integrated rate equation:

ln(6 x 10^-3 / 6 x 10^-5) = kt

ln(100) = kt

ln(100) ≈ 4.605

Now, substituting k = 10⁷ mol/L/s:

4.605 = (10⁷)t

t = 4.605 / (10⁷) = 4.605 x 10^-8 seconds

Choosing the Closest Option

From our calculation, the time it takes for the H⁺ concentration to drop significantly is approximately 4.605 x 10^-8 seconds. Among the options provided:

• 1) 6 x 10^-3 sec
• 2) 6 x 10^-6 sec
• 3) 6 x 10^-10 sec
• 4) 6 x 10^-9 sec

The closest value to our calculated time is option 4) 6 x 10^-9 seconds, which is the best choice based on the options given.