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Chapter 11: Problem 15

What volume of 0.25 M HCl solution must be diluted to prepare 1.00 L of 0.040 M HCl?

Short Answer

Expert verified
To prepare 1.00 L of 0.040 M HCl from a 0.25 M HCl solution, you need to dilute 160 mL of the 0.25 M HCl solution.

Step by step solution

01

Identify the given information

From the exercise, we know that: - Initial concentration (\(C_1\)) = 0.25 M - Final concentration (\(C_2\)) = 0.040 M - Final volume (\(V_2\)) = 1.00 L
02

Use the dilution formula to solve for the initial volume

Since we have all the values except for the initial volume (\(V_1\)), we can use the dilution formula: \(C_1 \times V_1 = C_2 \times V_2\) Plug in the given values: \(0.25 \times V_1 = 0.040 \times 1.00\)
03

Solve for the initial volume

Now, we need to solve for \(V_1\): \(V_1 = \frac{0.040 \times 1.00}{0.25}\) Divide and simplify: \(V_1 = 0.16\)
04

Convert the initial volume to the appropriate units

The initial volume is in liters, but it is more convenient to express it in milliliters. To convert, we can use the following conversion factor: 1 L = 1000 mL So, multiply the initial volume by 1000: \(V_1 = 0.16 \times 1,000\) Which gives: \(V_1 = 160\text{ mL}\) Now we know that \(160\) mL of 0.25 M HCl solution must be diluted to prepare 1.00 L of 0.040 M HCl.

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Most popular questions from this chapter

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