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30 September, 08:27

Part a a solution is made by mixing 9.00 mmol (millimoles) of ha and 3.00 mmol of the strong base. what is the resulting ph? express the ph numerically to two decimal places.

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  1. 30 September, 08:54
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    Assuming that the reaction is unimolecular on both reactant and product sides such that:

    HA + OH - - - > A - + H2O

    Therefore the amount of A - left is:

    mmol HA = 9.00 - 3.00 = 6.00

    mmol A - = 3.00

    pKa = - log Ka = 5.25

    Calculating for pH:

    pH = 5.25 + log 3.00/6.00

    pH = 4.95
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