Question

A 0. 300 m solution of hcl is prepared by adding some 1. 50 m hcl to a 500 ml volumetric flask and diluting to the mark with deionized water. What volume of 1. 50 m hcl must be added?.

Answer 1

you need to add 100 mL of the 1.50 M HCl solution to the 500 mL volumetric flask to create a 0.300 M HCl solution.

Step-by-step explanation:

To find out how much of the 1.50 M HCl solution needs to be added to the 500 mL volumetric flask to create a 0.300 M HCl solution, we can use the concept of dilution. The formula for dilution is:

C1V1 = C2V2

Where:

C1 = Initial concentration

V1 = Initial volume

C2 = Final concentration

V2 = Final volume

Given:

C1 = 1.50 M (concentration of the stock HCl solution)

V1 = volume of stock HCl solution to be added (unknown)

C2 = 0.300 M (desired final concentration)

V2 = 500 mL (volume of the volumetric flask)

Plugging in the values:

(1.50 M) * V1 = (0.300 M) * (500 mL)

Now, solve for V1:

V1 = (0.300 M * 500 mL) / 1.50 M

V1 = 100 mL

So, you need to add 100 mL of the 1.50 M HCl solution to the 500 mL volumetric flask to create a 0.300 M HCl solution.