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    Solubility and Titration

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    1. Tooth enamel consists of hydroxyapatite, Ca5(PO4)3OH (Ksp = 6.8x10-37). Fluoride ion added to drinking water reacts with Ca5(PO4)3OH to form the more tooth decay-resistant fluorapatite, Ca5(PO4)3F (Ksp =1.0x10-60). Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of Ca5(PO4)3OH and of Ca5(PO4)3F in water.

    2. A 35.00 mL solution of .2500 M HF is titrated with a standardized .1532 M solution of NaOH at 25◦C.
    (a) What is the pH of the HF solution before titrant is added?
    (b) How many milliliters of titrant are required to reach the equivalence point?
    (c) What is the pH at .50 mL before the equivalence point?
    (d) What is the pH at the equivalence point?
    (e) What is the pH at .50 mL after the equivalence point?
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    1. Tooth enamel consists of hydroxyapatite, Ca5(PO4)3OH (Ksp = 6.8×10-37). Fluoride ion added to drinking water reacts w/Ca5(PO4)3OH to form the more tooth decay-resistant fluorapatite, Ca5(PO4)3F (Ksp =1.0×10-60). Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of Ca5(PO4)3OH and of Ca5(PO4)3F in water.

    Solution:

    The balanced equation is:
    Ca5(PO4)3OH(s)  5 Ca2-(aq) + 3 PO43-(aq) + OH-(aq)

    So, the Ksp expression will be:

    Ca5(PO4)3OH  5 Ca2- + 3 PO43- + OH-
    5x 3x x

    The solubility of Ca5(PO4)3OH is 2.72 × 10-5 M.

    The balanced equation is:
    Ca5(PO4)3F(s)  5 Ca2-(aq) + 3 PO43-(aq) + F-(aq)

    So, the Ksp expression will be:

    Ca5(PO4)3OH  5 Ca2- + 3 PO43- + F-
    5x 3x x

    The solubility of Ca5(PO4)3F is 6.11 × 10-8 M.

    2. A 35.00 mL solution of .2500 M HF is titrated with a standardized .1532 M solution of NaOH at 25◦C.
    (a) What is the pH of the HF solution before titrant is added?
    (b) How ...

    Solution Summary

    This solution involves solving for solubility of compounds in water and various pH questions about titration.

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