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Chapter 4: Problem 73

Describe how to prepare \(1.00 \mathrm{~L}\) of \(0.646 \mathrm{M} \mathrm{HCl}\) solution, starting with a \(2.00 M \mathrm{HCl}\) solution.

Short Answer

Expert verified
To prepare \(1.00 \mathrm{~L}\) of a \(0.646 \mathrm{M}\) HCl solution from a \(2.00 \mathrm{M}\) HCl solution, one would need to take \(0.323 \mathrm{~L}\) of the concentrated solution and add \(0.677 \mathrm{~L}\) of water to it to achieve the target molarity.

Step by step solution

01

Identify Molarities and Desired Volume

From the problem, we know the molarity (\( M_1 \)) of our starting solution is \( 2.00 M \), and the molarity (\( M_2 \)) of our desired solution is \( 0.646 M \). We also know that we want to prepare \( 1.00 L \) of the diluted solution, that's \( V_2 \) = \( 1.00 L \).
02

Substitute Values into Dilution Formula

Substitute the known values into the dilution formula: \( (2.00 M)(V_1) = (0.646 M)(1.00 L) \).
03

Solve the Dilution Formula for V1

By rearranging the formula, we can solve for \( V_1 \), the volume of the concentrated solution needed. Divide both sides by \( 2.00 M \) to isolate \( V_1 \) on one side of the equation: \( V_1 = (0.646 M)(1.00 L) / 2.00 M = 0.323 L \). Therefore, to make \( 1.00 L \) of a \( 0.646 M \) solution from a \( 2.00 M \) solution, we need \( 0.323 L \) of the concentrated solution.
04

Dilute the Concentrated Solution

The next step will be to add enough water to the \( 0.323 L \) concentrated solution to make a \( 1.00 L \) solution. This would be \( 1.00 L - 0.323 L = 0.677 L \) of water. This makes a final volume of \( 1.00 L \) of \( 0.646 M \) HCl solution.

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