Consider the titration of H3C6H5O7 with NaOH. If it requires 0.00343 mol of NaOH

Question

Consider the titration of H3C6H5O7 with NaOH. If it requires 0.00343 mol of NaOH to reach the endpoint, and if we had originally placed 16.41 mL of H3C6H5O7 in the Erlenmeyer flask to be analyzed, what is the molarity of the original H3C6H5O7 solution? Consider the titration of H3C6H5O7 with NaOH. If it requires 0.00343 mol of NaOH to reach the endpoint, and if we had originally placed 16.41 mL of H3C6H5O7 in the Erlenmeyer flask to be analyzed, what is the molarity of the original H3C6H5O7 solution?

Explanation / Answer

H3C6H5O7 + 3NaOH --------------- Na3C6H5O7 +3 H2O

1 mole            3 mole

According to equation ,

for 3 moles of NaOH = 1 mole of H3C6H5O7

    0.00343 moles of NaOH= ?

                                      = 0.00343 x1/3 = 0.00114 moles fo citric acid

number of moles of citric acid = 0.00114 moles

volume of citric acid = 16.41 ml = 0.01641L

molarity = number of moles/volume in L

               = 0.00114/0.0164

               =0.0695 M

molarity of NaOH = 0.0695M

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