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Titrations

  • 09-04-2016 9:23pm
    #1
    100def
    Registered Users, Registered Users 2 Posts: 1 ✭


    Just wondering if someone could help me wrap my head around titrations, on the phys/chem course. Here's a question for reference.(its the calculation I have trouble with)


    A student titrated an ammonium hydroxide (NH4OH) solution against 25.0 cm3 volumes of a 0.10 M solution of hydrochloric acid (HCl). Ammonium hydroxide solution is a weak base. On average, 12.25 cm3 of the ammonium hydroxide solution was required for neutralisation. The equation for the titration reaction is as follows.
    NH4OH + HCl  NH4Cl + H2O
    ( a) Describe how a burette was prepared and filled for use in this titration. (12)
    ( b) Why is a conical flask usually preferable to a beaker as the container for a titration reaction? (6)
    ( c) Name a suitable indicator for this titration.
    State the colour change observed at the end point of the titration.
    Why should only one or two drops of indicator be used? (12)
    ( d) Calculate the concentration of the ammonium hydroxide solution in
    ( i) moles per litre,
    ( ii) grams per litre.


Comments

  • #2
    emersyn
    Registered Users, Registered Users 2 Posts: 430 ✭✭emersyn


    I have a feeling I may have gone wrong somewhere as I got a long decimal for the answer but here is the basics anyway. For all the titration calculations you need to use this formula. Slot in the values that you are given and then multiply it out to find the unknown value.
    So in the question you are asked for the concentration of the base in moles per litre, which is represented by M2 (or MB, depending on how you want to write it) in the formula. You get this by substituting in all the values and just multiplying it out until you get M2 = a number (about 0.204 in this case, I rounded up).
    Then to convert it to grams per litre, you need to multiply the molarity by the relative molecular mass. I wrote out a triangle that you can use to help remember how to convert between mass/molarity/relative molecular mass - cover up the one you're looking for and the two left will show you what to do. To get rel. mol. mass add together the atomic masses of all the elements in NH4OH, which is 35. Then multiply that by molarity, which you calculated in part (i). So 35 x 0.204 = 7.14 g/L.
    If that's an exam paper question you should have a look at the marking scheme for that year as they will usually have the work done out (and they will always give you the right answer unlike my wonky calculations :o)


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