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Chapter 4: Problem 76

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 \mathrm{M}\). How much water should you add? Assume volumes are additive.

Short Answer

Expert verified
You should add \(126.25 \mathrm{~mL}\) of water to dilute the HCl solution to exactly \(0.100 \mathrm{M}\).

Step by step solution

01

Identify known quantities

The problem provides the initial volume or \( V_1 = 505 \mathrm{~mL} \), and initial molarity or \( M_1 = 0.125 \mathrm{M} \). The final molarity \( M_2 = 0.100 \mathrm{M} \) after adding water is also given.
02

Use the formula for dilution

The formula for dilution is \( M_1 \cdot V_1 = M_2 \cdot V_2 \) where \( V_2 \) is the final volume of the solution after water is added.
03

Find the final volume

Rearrange the equation and solve for \( V_2 \) which is \( V_2 = \frac{(M_1 \cdot V_1)}{M_2} \). Substituting the given values, \( V_2 = \frac{(0.125 \mathrm{M} \cdot 505 \mathrm{~mL})}{0.100 \mathrm{M}} = 631.25 \mathrm{~mL} \).
04

Find the amount of water to add

The amount of water to add is simply the final volume minus the initial volume: \( 631.25 \mathrm{~mL} - 505 \mathrm{~mL} = 126.25 \mathrm{~mL} \).

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