24. A triprotic acid is titrated with 0.100 M NaoH. If 10.0 mL of a solution of

Question

24. A triprotic acid is titrated with 0.100 M NaoH. If 10.0 mL of a solution of the acid that is 5.00 g/200. mL is neutralized to a pale pink phenolphthalein endpoint by 20.00 ml of NaoH, what is the molar mass of the acid? A. 37.5 g/mol B. 3.75 x 103 g/mol C. 375 g/mol D. 7.50 x 102 g/mol E. 2.50 x 103 g/mol 25. Which of the following is used as a primary standard in the labs in which you determined the molar mass of an acid? A. Citric acid (Part C) B. A triprotic acid (Part C) C. A monoprotic acid (Part C) D. Oxalic acid (Part B) E. Sodium hydroxide (Parts B and C)

Explanation / Answer

24 )

moles of NaoH = molarity x volume (ml ) / 1000

moles of NaOH = 0.1 x 20 / 1000

moles of NaOH = 2 x 10-3

H3A + 3NaOH ---> Na3A + 3H20

from the reaction

moles of H3A = moles of NaOH / 3

moles of H3A = 2 x 10-3 / 3

moles of H3A = 0.66 x 10-3

given 10 ml of 5 g/ 200 ml acid

mass of acid = 10 x 5 / 200

mass of acid = 0.25 g

molar mass = mass / moles

molar mass of acid = 0.25 / 0.66 x 10-3

molar mass of acid = 375 g / mol

so the answer is option C

25 ) option D

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