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Melissa
Melissa asked in Science & MathematicsChemistry · 1 day ago

If a buffer solution is 0.170 M in a weak base ( 𝐾b=2.9×10−5) and 0.590 M in its conjugate acid?

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  • Robbie
    Lv 5
    1 day ago

    pOH = pKb + log ([conj.acid]/[Base])

    pOH = -logkb + log ([conj.acid]/[Base])

    pOH = 4.59 + log(0.590/0.17) = 5.13

    pH=14-pOH=14-5.13=8.87

  • hcbiochem
    Lv 7
    1 day ago

    For the equilibrium: B + H2O <--> BH+ + OH-

    Kb = 2.9X10^-5 = [BH+][OH-]/[B]

    2.9X10^-5 = (0.170) [OH-]/(0.590)

    [OH-] = 1.0X10^-4 M

    pOH = 4.00

    pH = 14.00 - pOH = 10.00

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