Question

Given a stock solution of 0.128 m bromide, solve for the bromide concentration obtained by diluting a 450 µl aliquot to the mark in a 25 ml volumetric flask.

Answer 1

Hey there!

Stock solution:

Concentracion bromide = 0.128 M

initial solution in volumetric flask =  450 µl

So , moles of bromide present:

450  µl in liters :

1 µl  =   = 1*10⁻⁶ liters

450 * ( 1*10⁻⁶ )  = 0.00045

0.128 * 0.00045 =>  57.6 * 10⁻⁶ moles

Now volume final is 25 mL , in liters : 0.025 L ou 25*10⁻³

so  new  bromide concentration:

57.6*10⁻⁶ / 25*10⁻³=> 2.304*10⁻³ M