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Homework Help: Acid Equilibrium

  1. Mar 6, 2006 #1
    Hydrocyanic acid is a weak acid (Ka = 4.9 x 10-10). If 0.1300 moles of gaseous HCN are dissolved in 0.8700 liters of water. Determine the pH of the HCN solution formed.

    HCN + H2O <-> H3O+ + CN-

    K_a = [CN-][H3O+]/[HCN]

    M HCN = 0.1300 mol HCN/0.87351338 L = 0.14882428 M HCN ??

    For L solution: 0.1300 mol HCN*(27.026 g/1 mol HCN) = 3.51338 g HCN

    870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L??

    K_a = [x][x]/[0.14882428 - x]

    Assuming 0.14882428 - x = 0.14882428,
    x = sqrt(4.9 x 10-10*0.14882428) = 8.539549E-6 = [H3O+]

    pH = -log(8.539549E-6) = 5.068565065 = 5.07 ???

    Thanks.
     
  2. jcsd
  3. Mar 6, 2006 #2

    GCT

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    set up seems fine to me
     
  4. Mar 7, 2006 #3

    Borek

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    Staff: Mentor

    You have only two significant digits in the Ka, thus you may safely assume concentration is 0.13/0.87M - and neglect volume change.

    Rest is OK.
     
  5. Mar 8, 2006 #4

    That doesnt make a big difference, but it is not right. Just assume total V is .870 L .
     
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