1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Amount of CaO required to removed HCO3- from 1 kg of hard water

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A sample of hard water contains 305 ppm of HCO3- ions. What is the minimum mass of CaO required to remove HCO3- completely from 1 kg of water sample?

    2. Relevant equations

    Ca 2+ + 2HCO3 + CaO = 2CaCO3 + H2O

    3. The attempt at a solution

    Since the HCO3- ions are 305 parts per million, it can be assumed that there are 0.305 g of the ions in 1 kg (or 1 L as the density is 1) of water. (I don't know if this assumption is correct)

    If I use the equation as above, then I get 0.1525 g of CaO which is not correct.

    I think that the whole trouble is with the ppm part. Can someone give me a clue?
  2. jcsd
  3. Aug 22, 2011 #2


    User Avatar

    Staff: Mentor

    Equation is incorrect - you are assuming there is some other source of Ca2+ to be present as well, while the only source of calcium is CaO.
  4. Aug 28, 2011 #3
    Sorry for the delayed response.

    Ca 2+ should not have been there, my mistake. The equation is

    CaO + 2HCO3- = CaCO3 + CO3-- + H2O

    Now one mole of CaO reacts with two moles of HCO3. This means that for half a mole of CaO, there ought to be one mole of HCO3.

    1 mole of HCO3 = 61 g and 1 mole of CaO is 56 g. In this case it would be 28 g of CaO.

    Therefore, 28/61 x .305 gives 0.140 g or 140 mg.

    Got it, thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook