0.415 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 0.1029 M NaOH. The initial burette reading is 1.81 mL; when the titration endpoint is reached, the final burette reading is 39.70 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.

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

answer: a chemical reaction

explanation:

im not wrong

the answer is 0.8m.

by: nicholas