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Ph of (buffer + HCl)

  1. Apr 14, 2007 #1
    Question

    A buffer solution is prepared by mixing 600.0 ml of 0.600 M HClO and 400.0 ml of 1.00 M NaClO. Calculate the pH of the solution after 40.0 ml of 3.00 M HCl is added to the buffer. For HClO, Ka = 3.0 X 10^-8.

    Work so Far

    Code (Text):

      A buffer solution os prepared by mixing 600.0 ml of 0.600 M HClO and
      400.0 ml of 1.00 M NaClO. Calculate the pH of the solution after
      40.0 ml of 3.00 M HCl is added to the buffer. For HClO,
      Ka = 3.0 X 10^-8.

      Work so Far:

      Initial Concentration of the Buffer Solution:


       Now NaClO dissolves to completion so we have an initial
       concentration of ClO:

                       1 M * .4 L
       [ClO]_initial: ------------- = .4 M
                       (.6 + .4)L


                        .6 M * .6 L
       [HClO]_initial: ------------- = .36 M
                        (.6 + .4)L


        HClO     <--->    H+ +    ClO-
        ----              --      ----

        .36               0       .4          initial
         -x               x        x          change
       .36-x              x      .4+x         equilibrium


                  x(.4+x)
        so Ka =  --------- ==> x = 1.206 X 10^-5
                  (.36-x)

        which is [H+] (its concentration) and for ClO-,
        its new concentration is .400012 in 1 Liter,
        thus there are .4 moles of ClO now.

      After HCl is added:

        Moles of HCl added: 3.00M * .04 L = .12 moles

        These .12 moles will react with the .4 moles of ClO
        and will be used up so there will be no change in
        the number of moles of H+.

        There is, however, a change in concentration. For if
        we have a concentration of 1.206 X 10^-5 M for H+,
        then our new concentration is:

                          -5
                1.206 X 10    * 1 L                -5
        [H+] = --------------------- = 1.15954 X 10
                (1 + .04) L

        pH = -log([H+]) = 4.936

     
    does this look right to you?
     
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 16, 2007 #2

    Borek

    User Avatar

    Staff: Mentor

    I am not sure if you are not overcomplicating things. These questions are usually solved simply assuming stoichiometrical reaction between strong acid and conjugated base present in the solution. This way you calculate amounts of acid and conujgated base and you put them into Henderson-Hasselnbalch equation. That's all.

    Borek
     
    Last edited by a moderator: Aug 13, 2013
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