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    hayzemay's Avatar
    hayzemay Posts: 7, Reputation: 1
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    #1

    Oct 17, 2007, 04:21 AM
    Relative molecular mass
    A dibasic organic acid (0.1208 g) requires 23.1 cm3 of 0.1 mol dm-3 NaOH for neutralisation. What is its approximate relative molecular mass?

    H2A + 2 NaOH à Na2A + 2 H2O
    templelane's Avatar
    templelane Posts: 1,177, Reputation: 227
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    #2

    Oct 17, 2007, 04:33 AM
    Another two parter

    1st work out the amount of moles of NaOH (see the other post/ the website I posted)

    Notice that H2A (that's your acid I'm assuming?) is in a ratio of 1:2
    So remember to times the NaOH molarity by two to get the moles of H2A

    Part two

    The MW (molecular mass) can be worked out using this equation

    mass (g) = amount (moles) x molecular weight (g/mole)

    just rearrange it to get the MW
    hayzemay's Avatar
    hayzemay Posts: 7, Reputation: 1
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    #3

    Oct 17, 2007, 04:51 AM
    so the moles for NaOH is 23.1/40 = 0.5775
    templelane's Avatar
    templelane Posts: 1,177, Reputation: 227
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    #4

    Oct 17, 2007, 06:36 AM
    0.1M (mol/dm^3) x (23.2cm^3/1000) = 0.00232M

    Right I have taken the concentration and times it by the volume -I have divided this by 1000 to get the units into dm^3 which is required to get the amount of moles (mol/dm^3) used to neutralise the acid

    Look at step three form here to see it written out clearer.
    Titration Calculations

    What you have done is taken the volume and multiplied it by the MW

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