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")#