Question

0.408 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an Erlenmeyer flask and is titrated with 0.1125 M NaOH. The initial burette reading is 0.82 mL; when the titration endpoint is reached, the final burette reading is 37.62 mL. How many moles of triprotic acid are neutralized during the titration? Provide your answer in decimal form (e.g. 0.123) to the correct number of significant figures.

Answer 1

Neutralization reaction is

H3PO4 + 3 NaOH .........> Na3PO4 + 3 H2O

one mole H3PO4 react with 3 mole NaOH.

volume of NaOH = (37.62 - 0.82) = 36.8 ml = 0.0368 L

thus

for 25 ml of  unknown triprotic acid solution,

mole of NaOH required = 0.0368 L * 0.1125 mole / L = 4.14 * 10^-3 mole.

thus

mole of unknown triprotic acid in 25 ml solution = 4.14 * 10^-3 / 3 = 1.38 * 10^-3 mole.

and

100 ml solution of unknown triprotic acid contains = 1.38 * 10^-3 mole * 100 ml / 25 ml = 5.52 * 10^-3 mole = 0.00552 mole.

thus

moles of triprotic acid are neutralized during the titration = 0.00552 mole.