Consider the titration of H 3 C 6 H 5 O 7 with NaOH. If it requires 0.00339 mol

Question

Consider the titration of H3C6H5O7 with NaOH. If it requires 0.00339 mol of NaOH to reach the endpoint, and if we had originally placed 17.83 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 0.00339 mol of NaOH to reach the endpoint, and if we had originally placed 17.83 mL of H3C6H5O7 in the Erlenmeyer flask to be analyzed, what is the molarity of the original H3C6H5O7 solution?

Explanation / Answer

H3C6H5O7    + 3 NaOH -----------------------> Na3 C6H5O7   + 3H2O

1mol                  3 mol

x                           0.00339

moles of H3C6H5O7 = 0.00339 /3 = 0.00113

molarity = moles / volume (Lit)

                = 0.00113 / (17.83 x 10^-3 )

                 = 0.063 M

molarity of the original H3C6H5O7 solution =0.063 M

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