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For DrBob222

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I'm not surprised the answer was wrong.
Chemical reactions are by mols and not by grams; therefore, 1 mol of each reactant produces 1 mol of each product but a mol is not a gram.
SO, convert 0.5839 g KHC2O4 to mols, convert mols KHC2O4 to mols KOH (that's the 1:1 part), then calculate the molarity of KOH. Post your work if you get stuck. I expect you will have no trouble.

when i converted KHC204 to mols, i got 0.00455 mols. when i divide that by the total volume (0.05699 L), i get 0.08, however, the answer is supposed to be 0.142.

  • For DrBob222 -

    Sorry I took so long to get back to you. It helps to post your work. Then we can catch the error. As it is I have no idea where the error is.
    mols KHC2O4 = 0.5839/128.126 = 0.004557.
    mols KHC2O4 = mols KOH
    M KOH = mols/L
    M KOH = mols KOH/Liters OF KOH (NOT TOTAL VOLUME). Now I know where the error is.
    M KOH = 0.004557/0.03199 = 0.14246 which rounds to 0.1425 to four significant figures.

    You may be thinking, "but the extra volume of other materials DID dilute the KOH " (and that is correct) but it also diluted the other materials at the same time. If you want to calculate it that way then change the molarities etc for all of the other components and the answer will come out to the same 0.1425. But the EASY way to look at the problem is this==and it's worth remembering. It isn't the volume that is reacting. It's the mols of one plus the moles of the other and you want to know the molarity at the BEGINNING of the titration and not at the end of the titration. I hope this helps.

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