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: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

answer; 4 electrons;