Creating a .0500 M HCl Calculation Help

[Bashyboy]

Hi, I have a problem regarding a lab question. It goes as follows:

A student required .0500 M solution of hydrochloric acid, HCl, and has on hand a .5000 M solution. Calculate how much more of the concentrated solution, in mL units, must be pipetted if 400 mL of dilute solution are required, and a 500 mL volumetric flask is the only one available.

Am I simply to use the M1 x V1 = M2 x V2 equation?

Thank you

 

[Borek]

Am I simply to use the M1 x V1 = M2 x V2 equation?

It will do. Beware - question is tricky. Not in terms of difficult calculations, more like a brain teaser.

Please ask HW and HW type questions in homework & coursework forums.

 

[symbolipoint]

The student "requires" 400 ml. of the resulting dilute solution, but because he only has a 500 ml. volumetric flask, he may choose to prepare 500 ml. of the dilute solution.

 

[Bashyboy]

Thank you both. And I am sorry; I didn't know there was a separate form for homework questions, thank you for letting me know.

 

[DeeNos]

for reaching out with your question. Yes, you are correct in using the M1V1 = M2V2 equation to solve this problem. Here is the breakdown of the calculation:

M1 = initial concentration of HCl solution (0.5000 M)
V1 = volume of initial solution to be used (unknown)
M2 = final concentration of HCl solution (0.0500 M)
V2 = total volume of final solution (400 mL + V1)

Using the equation, we can rearrange it to solve for V1:

V1 = (M2V2 - M1V1)/M1

Plugging in the given values, we get:

V1 = (0.0500 M)(500 mL) - (0.5000 M)(400 mL)/0.5000 M

Solving for V1, we get:

V1 = 40 mL

Therefore, the student will need to pipette an additional 40 mL of the concentrated solution into the 400 mL of dilute solution to achieve a final concentration of 0.0500 M. Keep in mind that this calculation assumes ideal conditions, so it is always important to double check your results and make any necessary adjustments in the lab. I hope this helps and good luck with your experiment!