If 0.900 g of oxalic acid, $H_2C_2O_4$ (90.04 g/mol) is completely neutralized with 0.500 M $LiOH$ , what volume of lithium hydroxide is required?

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Answer 1 · 2016-08-07T16:52:59

$sf(0.004color(white)(x)"L")$

Ethandioic acid is diprotic and reacts like this:

$sf(H_2C_2O_4+2LiOHrarrLi_2C_2O_4+2H_2O)$

This tells us that 1 mole $sf(H_2C_2O_4)-=$ 2 mole $sf(LiOH)$

The no. moles $sf(H_2C_2O_4)$ is given by:

$sf(n_(H_2C_2O_4)=m/M_(r)=0.900/90.04=0.001)$

So the no. moles of LiOH must be 2x this amount:

$sf(n_(LiOH)=2xx0.001=0.002)$

Concentration = no. moles solute/volume of solution i.e:

$sf(c_(LiOH)=n_(LiOH)/v_(LiOH)$

$:.$ $sf(v_(LiOH)=n_(LiOH)/c_(LiOH)=0.002/0.500=0.004color(white)(x)"L")$

If 0.900 g of oxalic acid, H_2C_2O_4H_2C_2O_4 (90.04 g/mol) is completely neutralized with 0.500 M LiOHLiOH, what volume of lithium hydroxide is required?