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...

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  • 4 years ago
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    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-]

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