In a clean 250-mL volumetric flask, 125 mL of a 6.0 M hydrochloric acid was added. This was then diluted with more water to reach the 250-mL mark. What is the molarity of the acid at this time?
use the equation CsVs=CdVd
Cs = Concentration of Stock (6.0M)
Vs = Volume of Stock (125mL)
Cd = Concentration of Diluted (??)
Vd = Volume of Diluted (250mL)
you are looking for the concentration or molarity of the diluted version of the hydrochloric acid because water was added to the 125mL of the acid.
so 6.0x125 = Cd x 250
no you isolate the Cd ..
6.0x125/250 = Cd
then you will get your answer, do you understand??
To find the molarity of the hydrochloric acid after dilution, we can use the equation:
M1V1 = M2V2
where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
In this case, the initial molarity (M1) is 6.0 M, the initial volume (V1) is 125 mL, the final volume (V2) is 250 mL, and we need to find the final molarity (M2).
Plugging these values into the equation, we have:
(6.0 M)(125 mL) = (M2)(250 mL)
To solve for M2, we can rearrange the equation as:
M2 = (6.0 M)(125 mL) / (250 mL)
M2 = 3.0 M
Therefore, the molarity of the hydrochloric acid after dilution is 3.0 M.