chemistry

Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%

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asked by Laura
  1. HNO2 ==> H^+ + NO2^-
    Initial:
    (HNO2) = 0.2
    (H^+) = 0
    (NO2^-) = 0

    Ka = (H^+)(NO2^-)/(HNO2)

    If 5.8% ionized, then after ionization:
    (H^+) = 0.2 x 0.058 = ??
    (NO2^-) = 0.2 x 0.058 = ??
    (HNO2) = 0.2 x (1-0.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100 - 5.8 = 94.2% OR 1.00 - 0.058 = 0.942).

    Plug these values into Ka expression above and solve for Ka.

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    posted by DBob222
  2. Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%


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    chemistry - DBob222, Saturday, April 5, 2008 at 4:50pm
    HNO2 ==> H^+ + NO2^-
    Initial:
    (HNO2) = 0.2
    (H^+) = 0
    (NO2^-) = 0

    Ka = (H^+)(NO2^-)/(HNO2)

    If 5.8% ionized, then after ionization:
    (H^+) = 0.2 x 0.058 = ??
    (NO2^-) = 0.2 x 0.058 = ??
    (HNO2) = 0.2 x (1-0.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100 - 5.8 = 94.2% OR 1.00 - 0.058 = 0.942).

    Plug these values into Ka expression above and solve for Ka.

    Here is my work:

    Ka = (H^+)(NO2^-)/(HNO2)

    = (0.0116)(0.0116) / (0.1884)

    = 1.34 X 10^-4 / (0.1884)

    = 7.14 X 10^-4

    My answer is still wrong:( Is there a step I missed?

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    posted by Sarah

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