Ask questions and get helpful answers.

A 1.575 g sample of ethanedioic acid crystals, H2C2O4. xH20 was dissolved in water and made up to 250cm^3 . One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm^3 of this solution of acid reacted with exactly 15.6 cm^3 of 0.160 mol/ dm^3 NaOH. Calculate the value of x

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
4 answers
  1. What exactly is x?

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    👤
    bobpursley
  2. H2C2O4. x H20

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. H2C2O4 + 2NaOH ==> Na2CO2O4 + 2H2O
    1.575 g H2C2O4.xH2O in 250 mL and 25 cc of that makes the sample titrated 1.575 x (25/250) = 0.1575grams.

    mols NaOH = M x L = approx 0.0025 but you need to redo all of this with better numbers.
    Convert mols NaOH to mols H2C2O4.xH2O. 1/2 * 0.0025 = approx 0.0012
    mols H2C2O4.xH2O = grams/molar mass or
    molar mass = g/mol = approx 0.16/0.00125 = approx 128 for H2C2O4.xH2O. You know H2C2O4 is 90 so xH2O must be approx 128-90 = about 38.
    38/molar mass H2O = 38/18 = about 2 when rounded to a whole number so x must be 2.
    If you redo the problem without the estimates you will get x = 2.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  4. What is the % of carbon in ethanedioc acid crystals
    C2H2O4.2H2O

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.