# How many mL of 3.00 M HCl are needed to prepare 600 mL of 0.400 M HCl solution?

## Question:

How many mL of {eq}3.00 \; M {/eq} {eq}HCl {/eq} are needed to prepare 600 mL of {eq}0.400 \; M {/eq} {eq}HCl {/eq} solution?

## Dilution of Solution:

A solution is a homogeneous mixture prepared with certain concentrations of one or more solute species present inside. A less concentrated version of a solution can be prepared through a procedure known as dilution. A sample of the starting solution is transferred to another container. To this sample is added more of the liquid solvent, but without increasing the solute quantities present. If each solute concentration is expressed as a molarity, then they will all decrease in this new diluted volume for the sample.

This is an aqueous solution where HCl is the solute. The 3.00 M solution is the stock and the 0.400 M is the diluted version. We need the HCl moles required for the diluted solution:

• Volume of diluted solution (given) = {eq}V_1 = 0.600 \ L {/eq}
• Molarity of HCl in diluted solution (given) = {eq}C_1 = 0.400 \ M {/eq}
• Moles of HCl = {eq}n = V_1 \times C_1 = 0.600 \ L \times 0.400 \ M = 0.240 \ mol {/eq}

Dividing by the stock HCl molarity gives the stock volume required:

• Stock HCl molarity (given) = {eq}C_2 = 3.00 \ M {/eq}
• Stock solution volume required = {eq}\dfrac{n}{C_2} = 0.240 \ mol \times \dfrac{L}{3.00 \ mol} = 0.0800 \ L = \boxed{80.0 \ mL }{/eq}

Calculating Dilution of Solutions

from

Chapter 8 / Lesson 5
69K

Learn what a solution is and how to properly dilute a new solution from a stock solution. Learn the dilution equation that combines molarity, the volume of stock solution and desired solution to determine how much stock solution is needed for the new solution.