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29 May, 10:22

You have 75.0 ml of 0.10 m ha. after adding 30.0 ml of 0.10 m naoh, the ph is 5.50. what is the ka value of ha?

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  1. 29 May, 12:18
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    First, we need to get moles of HA = molarity * volume

    = 0.1 m * 0.075 L = 0.0075 moles

    moles of NaOH = molarity * volume

    = 0.1 * 0.03 L = 0.003 moles

    from the reaction equation:

    HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)

    that means the final moles' HA = 0.0075 - 0.003 = 0.0045 moles

    when the total volume is = 0.075 + 0.03 L = 0.105 L

    ∴ [HA] = moles / volume

    = 0.0045 / 0.105 L = 0.043 m

    [A^-] = 0.003 / 0.105 L = 0.029 m

    then by using H-H equation:

    PH = Pka + ㏒[A^-] / [HA]

    by substitution, we can get Pka:

    5.5 = Pka + ㏒ (0.029 / 0.043)

    ∴ Pka = 5.67

    when Pka = - ㏒Ka

    5.67 = - ㏒ Ka

    ∴Ka = 2 x 10^-6
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