Calculate pH of Buffer Solution After HCl Added

[lizzyb]

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

  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?

 

[Borek]

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