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Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

Can someone please help show how to do these?

For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH.

2.1×10−4 M KOH

4.9×10−4 M Ca(OH)2

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  • david
    Lv 7
    2 months ago

    2.1×10−4 M KOH

      KOH  -->  K+  +  OH-  <<<  notice 1 KH forms 1 OH- ... same numbers

      2.1×10−4 M KOH   forms  2.1×10−4 M [OH -]

      [H3O+][OH-]  =  1x10^-14

       [H3O+]  =  4.76x10^-11 M

    pH = -log[H3O+]  = 10.32

    pOH = -log[OH-]  =  3.68

    ====================================4.9×10−4 M Ca(OH)2

      Ca(OH)2  -->  Ca2+  +  2OH-   <<<<  OH- is 2 times  Ca)OH)2

      [OH-]  =  9.8x10^-4 M

      [H30+]  =  1x10^-14 / 9.8x10^-4  =  1.02x10^-11 M

    pH = 10.99

    pOH = 3.01

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  • 2 months ago

    Both calculations go exactly the same way as in

    https://answers.yahoo.com/question/index?qid=20200...

    A)

    [OH−] = 2.1×10^−4 M

    [H3O+] = 10^-14 / (2.1×10^−4) = 4.8×10^−11

    pH = - log (4.8×10^−11) =  10.3

    pOH = - log (2.1×10^−4) =  3.7

    B)

    Ca(OH)2 → Ca{2+} + 2 OH{-}

    Supposing complete dissociation:

    [OH-] = 2 x (4.9×10^−4 M) = 9.8×10^−2 M

    [H3O+] = 10^−14 / (9.8×10^−2) = 1.02×10^−13 M

    pH = - log (1.02×10^−13) = 13.0

    pOH = - log (9.8×10^−2) = 1.0

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