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21 September, 16:29

A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium hydroxide solution is added? the ka of hydrazoic acid is 1.9*10-5.

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  1. 21 September, 18:03
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    First, we need to calculate moles of hydrazoic acid NH3:

    moles NH3 = molarity * volume

    = 0.15 m * 0.025 L

    = 0.00375 moles

    moles NaOH = molarity * volume

    = 0.15 m * 0.015 L

    = 0.00225 moles

    after that we shoul get the total volume = 0.025L + 0.015L

    = 0.04 L

    So we can get the concentration of NH3 & NaOH by:

    ∴[NH3] = moles NH3 / total volume

    = 0.00375 moles / 0.04 L

    = 0.09375 M

    ∴[NaOH] = moles NaOH / total volume

    = 0.00225 moles / 0.04 L

    = 0.05625 M

    then, when we have the value of Ka of NH3 so we can get the Pka value from:

    Pka = - ㏒Ka

    = - ㏒ 1.9 x10^-5

    = 4.7

    finally, by using H-H equation we can get PH:

    PH = Pka + ㏒[salt / basic]

    PH = 4.7 + ㏒[0.05625/0.09375]

    ∴ PH = 4.48
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