Question

Consider the titration of H3C6H5O7 with NaOH. Let’s say that you need to add 0.0387 moles of NaOH in order for the indicator dye to change color. How many moles of H3C6H5O7 must you have had to start? Report answer to the correct sig figs. Do not provide units.

Answer 1

Citic acid (H3C6H5O7) is a triprotic acid, therefore 3 moles of NaOH are required to neutralise one mole of H3C6H5O7. At the point of neutralisation the indicator changes color.

So we have

3 moles of NaOH = 1 mole of H3C6H5O7.

1 moles of NaOH = 1/3 moles of H3C6H5O7 (one mole of NaOH will neutralise 1/3 moles of H3C6H5O7)

And 0.0387 moles of NaOH (Moles given in question) will neutralise (1 / 3) X 0.0387 = 0.0129 moles of H3C6H5O7

Hence the moles of H3C6H5O7 initialy present must be equal to = 0.0129