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What would be the proper way to solve a titration problem?

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    "A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added and the resulting solution is titrated with 2.50 M HCl (aq) solution. The indicator changes colour signaling that the equivalence point has been reached after 17.0 mL of the hydrochloric acid solution has been added."

    a) What is the molar mass of the metal hydroxide?
    b) What is the identity of the alkali metal cation: Li+, Na+, K+, Rb+, or Cs+?

    2. Relevant equations
    Molarity = moles solute / volume of solution in liters

    3. The attempt at a solution
    First, I found the moles for HCl:
    moles solute / volume of solution in liters = molarity
    moles solute = molarity (volume of solution in liters)
    0.0425 mol HCl = 2.5(0.017)
    Now I guessed that 0.0425 mol of HCl is supposed to react with 0.425 (?)OH to produce H2O and (?)Cl. Then I tried to find the moles of all the possible alkali metal hydroxides by adding the molar mass of hydrogen and oxygen, then adding them to the different alkali metals. Then I took my given mass of 4.36 g and divided it by each of those molar masses. If the result was not 0.0425, then I eliminated it as a solution. This is my final result:
    4.36/102.47514 = 0.0425 mol RbOH, where 102.47514 = 1.00794 + 15.9994 + 85.4678. I then concluded that the alkali metal was rubidium, seeing how the numbers matched up.
    However, I am completely sure this is not the correct way to do this problem. Could someone tell me how they would do this titration problem?
     
  2. jcsd
  3. Feb 20, 2015 #2

    Quantum Defect

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    I think your reasoning looks fine. What I would say is that you know that the moles of hydroxide equals moles of acid. (All alkali metals have +1 charge as ions). You also know grams of alkali hydroxide. Molecular weight is grams divided by moles. So the molecular weight is 4.36 g/ 0.0425 mol = 103 g/mol. Since RbOH is closest to this, the sample is most likely RbOH.
     
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