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

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

each chemical family is represented as a group or column on the periodic table. based on chemical properties, there are 6 types of chemical families. group 1 represents alkali metals, group 2 represents alkaline earth metals, groups 3 to 12 represent transition metals, groups 13 to 16 represent non-metals and other metals, group 17 represents halogens and group 18 represents noble gases.