Problem set and Chapter 2 material

1.  Calculate pH of a solution when 100ml of 0.1M NaOH is added to 150 ml of 0.2 M acetic acid.  (pka = 4.76)

            0.1 l x 0.1 moles/l = 0.01 moles NaOH
                0.15 l x 0.2 moles/l = 0.03 moles acetic acid

                        pH = 4.76 + log 0.01/0.03-0.01

                        pH = 4.46
 
 

2.  Calculate concentration of acetic acid and sodium acetate needed to make a buffer of pH 5.0 with a total acetate
concentration of 0.2 M. (pka = 4.76)

            K = (H) (A)             1.75 x 10-5 = (1 x 10-5) (X)
                         (HA)                                           (0.2 - X)
 

                X = 0.13M = A
                        0.07M = HA

3.  Calculate how to make 150 ml of a 0.125M formate buffer, pH 3.5 from 1.0M formic acid and 1.0 M NaOH.
(pka = 3.75)
 
            Need 0.125M formic acid because it contributes all the formate to the buffer.  Thus,

                            (0.125 moles/l) (0.150 l)  =  (1 mole/l) (X);    X= 18.75 ml formic acid
 

            Next calculate using either the H-H equation or the K equation, how much base you need.

                            1.78 x 10-4 + 3.12 x 10 -4 (X) / (0.125 - X)    X = 0.045M

                            (0.045 moles/l) (0.150 l) = (1 mole/l) (X);   X = 6.75 ml of NaOH

            Add the two components and bring volume up to 150 ml.

4.  Calculate how to make 8 liters of a 0.1M phosphate buffer, pH 12.1 from solid K2HPO4 (mw = 174.18 g/mole;  pka = 12.67) and your choice of 5 M HCL or 5 M KOH.

            A similar strategy to problem 3 is used in that all the phosphate component comes from K2HPO4.

                The answer:  139.3 g K2HPO4 + 33.9 ml KOH + water to 8 liters.
 

Voet and Voet problem 8.

            For the following calculations all are either strong acids or bases thus they totally dissociate

    a.  0.1 M HCl = pH 1.0

    b.  0.1 M NaOH = pH 13

    c.  3 x 10-5 M HNO3 = pH 4.52
 

Voet and Voet problem 15.

                For the following problem you must calculate the concentration at all pHs.

        Therefore at pH 9.0, 10.0 and 9.4 for a 0.1 M buffer

        9.0 = 9.25 + log X/0.1-X     X = 0.36M

        10.0 = 9.25 + log X/0.1-X   X = 0.085M

        9.4 = 9.25 + log X/0.1-X     X = 0.059M

            Let a = the fraction of pH 9.0 buffer required, then

                0.036M (a) + 0.085M (1-a) = 0.059M   a = 0.53

            Thus,

                0.53 x 200ml = 106 ml of the pH 9.0 buffer + 94 ml of the pH 10 buffer gives the desired solution