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March 29, 2017

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Consider the titration of 25.0mL of 0.10M HAc with 0.10M NaOH. That is, NaOH is added to HAc. (a)pH at the beginning of titration. (b)pH at the equivalence point of the titration. (c) pH at the midpoint of the titration.

  • Chemistry - ,

    a.) pH=-log[H+]=-log(0.10M HAc)

    b.) At the 1/2 eq. pt., 1/2 of the HAc will react with NaOH. So, 25 z 10^-3L (0.01M)= moles of HAc

    1/2*moles of HAc= moles of unreacted HAc

    moles of unreacted HAc/total volume (37.5mL)= M of unreacted HAc

    pH=-log[H+]=-log[M of unreacted HAc]
    c.) Neutralization reaction, pH= 7

  • Chemistry - ,

    I apologize,

    HAc + H20---> H30+ Ac

    HAc= acetic acid, which is a weak acid.

    a.)

    Ka=1.8 x 10–5=[x][x]/0.100M-x

    5% rule allows you to ignore -x


    Ka=1.8 x 10–5=x^2/0.100M


    solving for x,

    x=1.34 x 10^-3

    pH=-log(1.34 x 10^-3 M)

    B.)

    pKa=-log(Ka)

    pH=pKa+log([Ac/HAc])

    Concentrations of Ac and HAc are eqaul, pH=pKA.


    C.)
    Ac- + H2O ---> OH +HAc

    Kb=Kw/Ka=[OH][HAc]/[Ac-]

    Ac-=1/2*(0.10M)

    sqrt*(Kb[Ac-])=OH

    -log[OH]=pOH
    14-pOH=pH


    I apologize about that one; I assumed HAc was a strong acid. Oppps.

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