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Calculate [OH-] and pH for the following strong base solution: 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL

Question ID
533142

Created
April 18, 2011 11:52pm UTC

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0

URL
https://questions.llc/questions/533142

Answers
3

Views
2920

3 answers

  1. [Ca(OH)2] = 0.0105M; therefore, 10 mL diluted to 500 must have a concn of
    0.0105M x (10/500) = ??M
    Since there are two OH ions per molecule of Ca(OH)2, (OH^-) must be twice that, then pOH = -log(OH^-).
    pH + pOH = pKw = 14; use that to calculate H+.

    Answer ID
    533224

    Created
    April 19, 2011 1:52am UTC

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    0

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  2. Thank you so much! I would really appreciate it!
    [OH-]= 4.2x10^-4 M
    pH=10.6 or 11
    Are these answers correct?

    Answer ID
    533289

    Created
    April 19, 2011 3:19am UTC

    Rating
    -1

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  3. Yes, that's correct but I wouldn't round it to 11 but keep it as 10.62. (10.62 has two significant figures --besides the 10 which comes from the log part--and you are allowed two--and I would keep 10.62.

    Answer ID
    533322

    Created
    April 19, 2011 4:46am UTC

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    0

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