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

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

= enthaply

= energy of products

= energy of reactants

as , thus comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

the answer is definitely c.