Calculate the pH of Buffer Solution with Ka=1.34x10^-5

[lizzyb]

Question

A buffer solution is prepared by mixing 380.0 ml of 0.250 M propionic acid, C2H5COOH, and 120.0 ml of 0.350 M sodium propionate, C2H5COONa. Calculate the pH of the prepared buffer solution. For propionic acid, Ka = 1.34 X 10^-5.

Work thus far

  We have this

  C2H5COOH <--> C2H5COO- + H+

  Since the C2H5COONa goes to completion and is left
  with C2H5COO-, we can determine the initial amount
  of C2H5COO-, and then we use the x-method to determine
  the pH.

  Initial Molarities:

                 .250 M .380 L     .250 * .380
   [C2H5COOH] = --------------- = ------------ = .19 M
                 (.380 + .120)L        .5      


                 .350 M * .12 L     .350 * .12
   [C2H5COO-] = ---------------- = ------------ = .084 M
                 (.380 + .120) L      .5

  C2H5COOH   <-->     C2H5COO-    +        H+

    .19                 .084               0       initial
     -x                  x                 x       change
   .19-x               .084+x              x       equilibrium

  so,

                         (.084 + x) x
     Ka = 1.34 X 10^-5 = --------------
                          (.19-x)

     ==> x = 3.029 X 10^-5, pH = -log(x) = 3.002

does this look okay to you?

 

[symbolipoint]

Yes. ... ..

 

[lizzyb]

thank you! :-)