Calculate the ph at the following points in a titration of 40 ml (0.040 l) of 0.100 m barbituric acid (k_a = 9.8 times 10^-5) with 0.100 m koh. (a) no koh added (b) 20 ml of koh solution added (c) 39 ml of koh solution added (d) 40 ml of koh solution added (e) 41 ml of koh solution added

a.pH = 2.51  

b.pH = 4.01

c.pH = 5.60

d.pH = 8.35

e.pH = 11.09

Explanation:

Ka = 9.8 x 10^-5

Ka = 4.01

a) no KOH added :

pH = 1/2 (pKa - log C)

    = 1/2 (4.01 - log 0.1)

pH = 2.51  

b)

millimoles of acid = 40 x 0.1 = 4

millimoles of KOH = 20 x 0.1 = 2

this is half equivalence point :  

here pH = pKa

pH = 4.01

c)  

millimoles of KOH = 39 x 0.1 = 3.9

acid +   KOH > salt + H2O

4         3.9                         0  

0.1       0                         3.9  

pH = pKa + log [salt / acid]

    = 4.01 + log [3.9 / 0.1]

   = 5.60  

pH = 5.60

d)

Millimoles of KOH = 40 x 0.1 = 4

 Here, salt remains.

salt concentration = 4 / 40 + 40 = 0.05 M

pH = 7 + 1/2 (pKa + log C)

   = 7 + 1/2 (4.01 + log 0.05)

  = 8.35

pH = 8.35

e)

pH = 11.09

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