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posted by  Click here on 5/11/2008 11:36:38 AM  |  status: Closed  

pH question answer provided

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
How much Ca(OH)2 (molar mass = 74.1 g/mole) is needed to prepare 750 mL of solution with a pH of 10.80?

answer: 0.018g
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posted by richmondMike on 5/11/2008 12:03:22 PM  |  status: Live
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"thank you very much"
Response Details:
Since pH = 10.8, then pOH = 14.0 - 10.8 = 3.20.
∴[OH] = 10-3.20.
.
 Let the number of mole of Ca(OH)2 is x.
∴[Ca(OH)2] = (1/2)[OH-] = (1/2)10-3.20 = 0.000315M = x / 0.750L.
∴x = 0.000237mol.
Answer = (0.000237mol)(74.1 g/mol) = 0.018g.
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