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chemistry

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A weak base that encompasses a concentration of 1.4 mol per litre has a percent ionization of 0.63%. Calculate the Kb of this weak base

  • chemistry - ,

    Let's call a weak base BOH.
    BOH ==> B^+ + OH^-

    Kb = (B^+)(OH^-)/(BOH)

    Look at the ICE chart.
    I = initial concn (before ionization):
    BOH = 1.4 mols/L.
    (B^+) = 0
    (OH^-) = 0

    C = change in concn:
    (B+) = +1.4*0.0063
    (OH^-) = +1.4*0.0063
    (BOH^-) = 1.4*0.9937 (0.9937 is the fraction from 100%-0.63% = 99.37%)

    E = equilibrium concn:
    (B^+) = 1.4*0.0063 = ??
    (OH^-) = 1.4^0.0063 = ??
    (BOH) = 1.4-(1.4*0.0063) OR (1.4*0.9937= ??)

    Substitute the equilibrium numbers into Ka expression and calculate Ka.

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