Please calculate the [OH-] and [Ca2+] of the Ca(OH)2 solution...?

At equivalence point, 20 mL of 0.1 M of HCl was use to titrate 10 mL of Ca(OH)2 solution. (Show all work for full points)

Please calculate the [OH-] and [Ca2+] of the Ca(OH)2 solution .

PLEASE SHOW ALL OF THE WORK SO I UNDERSTAND HOW TO DO THIS.

I think I'm overthinking it too much...

Update:

Edit: Ksp = [Ca^2+][OH–]^2 is the equilibrium constant expression if that's needed btw.

2 Answers

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  • Fern
    Lv 7
    5 years ago
    Favorite Answer

    Ca(OH)2(aq) + 2HCl(aq) =⇒ CaCl2(aq) + 2H2O(l)

    0.10 moles HCl/1liter x 20 mL x 1 liter/1000 mL = 0.0020 moles HCl

    0.0020 moles HCl x 1 mole Ca(OH)2 / 2moles HCl = 0.0010 moles Ca(OH2

    Ca(OH)2 =⇒ Ca+2 + 2OH-

    Molarity of Ca(OH)2 = 0.0010 moles Ca(OH)2 / 10.0 mL x 1000 mL/1liter = 0.10 M

    0.10 Moles Ca(OH)2 /liter x 1 mole Ca+2/ 1mole Ca(OH)2 = 0.10 M = [Ca+2]

    0.10 moles Ca(OH)2 / liter x 1 mole OH- / 2 moles Ca(OH)2 = 0.20 M = [OH-]

  • 5 years ago

    On the final part of Fern s answer, shouldn t it be 2 mole OH/1 mol Ca(OH)2, rather than 1 mol OH/2 mol Ca(OH)2?

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