pH = -log 1 x 10-2

pH = 2

b) 0,77 mol hidrazin dalam 250 ml air

 [N2H4] = n/V = 0,77/0,25 = 3,08 M

[OH-] = √Kb . [N2H4]

= √1,3 x 10-6 . 3,08 M

= 2 x 10-3 M

pOH = -log 2 x 10-3 

pOH = 3 - log 2

pH = 14 - (3 - log 2)

pH = 11 + log 2

pH = 11 + 0,3

pH = 11,3

c) 0,976 gram asam benzoat dilarutkan ke dalam 500 ml air

Mm [C6H5COOH] = Mm [7 . C] + [6 . H] + [2 . O]