Ask Question
18 January, 23:10

A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is needed to titrate a 25 mL aliquot of the unknown monoprotic acid. From this info, calculate the molar mass of the unknown acid (g/mol).

+5
Answers (1)
  1. 18 January, 23:26
    0
    36.45 g/mol

    Explanation:

    The titration of a monoprotic acid with a base as NaOH is:

    HX + NaOH → NaX + H₂O.

    That means 1 mol of acid reacts per 1 mol of base.

    10.0mL of 0.10M NaOH are:

    0.0100L ₓ (0.10mol / L) = 0.0010 moles NaOH ≡ 0.0010 moles HX

    If 72.9mg of the monoprotic acid are dissolved in 50.0 mL of water, 25 mL of the solution will contains:

    72.9mg * (25mL / 50mL) = 36.45mg = 0.03645g of HX.

    Molar mass is ratio between mass and moles of substance, that is:

    0.03645g of HX / 0.0010 moles HX = 36.45 g/mol
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers