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hayzemay · Oct 17, 2007, 04:21 AM

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 · 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 · Oct 17, 2007, 04:51 AM

so the moles for NaOH is 23.1/40 = 0.5775

templelane · 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