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13 December, 16:01

Calculate the change in ph when 3.00 ml of 0.100 m hcl (aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3 (aq) and 0.100 m in nh4cl (aq). a list of ionization constants can be found here.

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  1. 13 December, 18:44
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    The change in pH is calculated by:

    pOH = Protein kinase B + log [NH4+] / [NH3]

    Protein kinase B of ammonia = 4.74

    initial potential of oxygen hydroxide = 4.74 + log 0.100/0.100 = 4.74

    pH = 14 - 4.74=9.26

    moles NH4 + = moles NH3 = 0.100 L x 0.100 M = 0.0100

    moles H + added = 3.00 x 10^-3 L x 0.100 M=0.000300

    NH3 + H + = NH4+

    moles NH3 = 0.0100 - 0.000300=0.00970

    moles NH4 + = 0.0100 + 0.000300=0.0103

    pOH = 4.74 + log 0.0103 / 0.00970 = 4.77

    oH = 14 - 4.77 = 9.23

    the change is = 9.26 - 9.23 = 0.03
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