Solubility for Ca(OH)2?

I had a chem lab that I can't figure out. I am titrating Ca(OH)2 with 0.05674 M HCl. My volume of HCl added was 10.51mL. I am supposes to find the solubiliy of Ca(OH)2 and then it's Ksp. How do I do this.

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  • 1 decade ago
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    I assume that you got a saturated solution of Ca(OH)2 and then took an aliquot and titrated this

    Equation

    Ca(OH)2 + 2HCl = CaCl2 + 2H2O

    Moles HCl = conc * vol

    = 0.05674 * 0.01051

    = 5.963*10^-4

    So one mole of Ca(OH)2 will react with 2 moles of HCl; so 5.963*10^-4/2= 2.982*10^-4 moles of Ca(OH)2 will react with 5.963*10^-4 moles of HCl

    Ca(OH)2 = Ca2+ + 2OH-

    1 mole of Ca(OH)2 will produce 1 mole of Ca2+ ions and 2 moles of OH- ions; so 2.982*10^-4 moles of Ca(OH)2 will produce 2.982*10^-4 moles of Ca2+ and 5.963*10^-4 moles of OH-.

    Ksp = [Ca2+][OH-]^2

    That is as far as I can get with the info that you have given me.

    You need to work out the concentration by dividing the moles of Ca2+ and OH- by the aliquot volume taken. Plug these into the Ksp formula and you have your answer

  • 1 decade ago

    Balenced equation for the reaction, if I did it correctly, is:

    Ca(OH)2 + 2HCl -> CaCl2 + H2O

    So 2 moles of HCl are required to react with one more of Ca(OH)2.

    So the math is:

    (0.01051 L of HCl) x (0.05674 moles HCl)/1L x (1 mole Ca(OH)2)/(2 moles HCl) = (moles Ca(OH)2)

    Solving for that would equal the number of moles of Ca(OH)2 in the solution:

    (moles Ca(OH)2)/(0.01051L) = molarity of Ca(OH)2

    Ksp is the concentrations of the product raised to their coefficients devided by the concentrations of the reactions raised to their coefficients, so:

    Ksp = ([CaCl2][H2O])/([Ca(OH)2][HCl]2), where the brackets denote the concentration (obviously lol).

    You know the concentration of of HCl and Ca(OH)2, and from the math above, you know how to find the concentration of CaCl2 and H2O. Good luck!

    -Andrew

    Source(s): Chemistry major at UIC
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