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Titration exam question

  1. May 6, 2017 #1
    1. The problem statement, "all variables and given/known data
    A solution of nitric acid, HNO3, of concentration 100 g dm-3, can be used to artificially age wood.

    A sample of nitric acid was tested by pipetting 10.00 cm3 of the acid into a 250 cm3 volumetric flask and filled up with deionised water to the mark. A titration took place where 19.95 cm3 (mean titre) of sodium hydroxide solution with a concentration of 0.0800 mol dm-3 was added to 25 cm3

    Calculate the concentration of the undiluted nitric acid in g dm-3

    2. Relevant equations
    MaVa = MbVb

    3. The attempt at a solution
    Using the above equation I got the concentration of the 25 cm3 of nitric acid solution as 0.6384mol dm-3, which is 40.2192g dm-3, but how do I find the undiluted concentration?
     
  2. jcsd
  3. May 6, 2017 #2

    epenguin

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    Well firstly you will need to get into the habit for all quantitative work of checking automatically for order-of-magnitude reasonableness, so some mistakes hit you in the eye. You've taken some almost 0.1 M NaOH, you've neutralised it with a roughly equal volume of acid, and you say the acid is 6× as concentrated! It's a decimal point error, but you home in on detecting it faster by the above type of reasoning than by going through the calculations in detail. And if this is happening in the last 30 seconds of an exam you just write "this figure is unreasonable, about an order of magnitude too high" and you would in this case get most of the credit, at least if it's humans marking.

    After you've done that you have the concentration of your diluted nitric acid. You had diluted this 25× and you ask what was the original concentration?
     
  4. May 6, 2017 #3
    Just redid the equation and got values ten times smaller than before, so its 4.02192 g dm-3 for the 25 cm3 for the acid used. So, x25 now?
     
  5. May 6, 2017 #4

    epenguin

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