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How much water must be added to 500.mL of .200M HCL to produce a .150M solution?

(.500mL x .200M)/.150M = .667mL

A chemistry student needs 125mL f .150M NaOH solution for her experiment, but the only solution avail. is 3.00M. Describe how the student could prepare the solution needed.

I don't know about this one

  • Chem -

    For the first one, check your work. I think you have an error of about 1000 times. I think 500*0.200/0.150 = 667 mL. However, that is the volume you will have, not how much water must be added. Water that must be added is 667-500 = ??

    For the second one,
    M*mL = M*mL.
    125 mL * 0.15 M = 3.0 M x mL NaOH.
    mL 3.0 M NaOH = (125*0.15/3.0) = 6.25 mL of the 3.0 M NaOH.
    So measure exactly 6.25 mL of the 3.0 M NaOH and make to exactly 125 mL. (All of that may be easier said than done. Measuring exactly 6.25 mL could be done rather accurately with a buret (not many pipets of 6.25 mL) BUT making to exactly 125 mL would be hard to do for there aren't any 125 mL volumetric flasks. At least I've not seen one. But there could be a way around that. The student could add 12.50 mL of the 3.0 M NaOH to a 250 mL volumetric flask, then add water to the mark. That would be exact.

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