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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)

  • Chemistry -

    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.

  • Chemistry -

    How were you able to get the 1.6?

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