A .307 gram smaple of an unknown triprotic acid is titrated to the third equivalence point using... 35.2 mL of .106 M(molar) NaOH. Calculate the Molar Mass of the acid

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jerichorayel | College Teacher | (Level 2) Senior Educator

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Hello! :)


To answer this problem, we need to write a chemical equation for the titration of the triprotic acid.

we let the triprotic acid be H3A  

H3A + NaOH    --->  H2A- +  Na+  + H2O

H2A- + NaOH   ---> HA2- + Na+ + H2O

HA2- + NaOH  ---> XA- +  Na+  + H2O


H3A  + 3NaOH  ----> Na3A + 3H2O     this will be our working  



next we need to get the moles of NaOH

0.106 moles/L x 0.0352L =

=0.0037312 moles NaOH = 3.7312x10^-3 moles


From the equivalent of NaOH we can derive the equivalents of H3A using the balanced equation above (working equation).

since it is a triprotic, 

3.7312x10^-3 moles NaOH x ( 1moles H3A / 3 molesNaOH ) =


                      =1.24437x10^-3 moles H3A

*In the reaction above(working equation), the reaction needs 3 moles of NaOH in order to reach the third equivalence point. 

Now we can get the molecular weight of the acid

molecular weight = mass/moles

                         = 0.307 grams X /1.24437x10^-3 moles H3A

                         = 246.84 g/mole



                         = 247 grams/mole  is the answer..

hope this helps :)


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