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25 February, 15:42

Be sure to answer all parts. a buffer is prepared by mixing 50.0 ml of 0.050 m sodium bicarbonate, nahco3, and 10.7 ml of 0.10 m naoh. (ka = 4.7 * 10-11) what is the ph?

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  1. 25 February, 18:33
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    Moles HCO3 - = 0.0500L*0.050M = 0.0025

    Moles OH - added = 0.0107L*0.10M = 0.00107

    Moles CO32 - forms = 0.00107

    Moles HCO3 - in excess = 0.0025-0.00107=0.00143

    Total volume = 10.7 + 50.0 = 60.7mL = 0.0607L

    [HCO3-] = 0.00143 / 0.607 = 0.0236M

    [CO32-] = 0.00107/0.0607 = 0.0176M

    pKa = 10.3

    PH = 10.3 + log 0.0176/0.0236 = 10.2.
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