Asked by Emma
Prepare a 0.0200 mM dye solution at low pH by dilution 0.40 ml of the stock 0.100 mM chemical X into the appropriate volume of 2.5 M HCl.
What is the amount of HCL required for dilution, the final concentration of HCl in the solution, and the pH?
I know there's the MV = MV, but I'm not sure how to begin to solve this. The 2.5 M HCl throws me off since there's two concentrations now (2.5 and the .02)
What is the amount of HCL required for dilution, the final concentration of HCl in the solution, and the pH?
I know there's the MV = MV, but I'm not sure how to begin to solve this. The 2.5 M HCl throws me off since there's two concentrations now (2.5 and the .02)
Answers
Answered by
DrBob222
The 2.5 M HCl is the diluting agent and has nothing to do with the calculation for the dye. (Another way of saying it is that the 2.5 M HCl is just water as far as the dye dilution is concerned.). My MV = MV gets 2 mL.
The HCl is 2.5 x (1.6/2.0) = ?? if we assume the volumes are additive.
The HCl is 2.5 x (1.6/2.0) = ?? if we assume the volumes are additive.
Answered by
Emma
How were you able to get the 1.6?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.