Question

I need to find the M of citric acid from my lab experiment. (This is a common titration experiment with NaOH solution and citric acid soda.

H3C6H5O7 (aq) +3NaOH (aq) ---> Na3C6H5O7 (aq) + 3H2O (l)

The average molarity of my NaOH solution is 0.00896. My first trial resulted in a net 13.26mL. Molar mass of citric acid is 192.12352. How do I figure Mcitirc acid? Mol/can (355 mL) and g citric acid/can?

Answer 1

H3C6H5O7 (aq) +3NaOH (aq) ---> Na3C6H5O7 (aq) + 3H2O (l)

no of moles of NaOH = molarity * volume in L

                                   = 0.00896*0.01326 = 0.00012 moles

from balanced equation

2 moles of NaOH react with 1 mole of H3C6H5O7

0.00012 moles of NaOH react with = 1*0.00012/2   = 6*10^-5 moles of H3C6H5O7

molarity of H3C6H5O7   = no of moles/volume in L

                                        = 6*10^-5/0.355   = 0.00017M >>>>>answer