Tentukan pH dari
a. H2SO4 0,001 M
jawab : [H+] = a . M = 2 . 10^-3 = 2.10^-3
pH = - log H+ = - log 2.10^-3 = 3 - log 2
b. 3,65 gram HCl (Ar H =1, Ar Cl = 35,5 ) dilarutkan dalam 500 ml air
jawab : mol = massa/Mr = 3,65 / 36,5 = 0,1
M = mol / volume = 0,1 / 0,5 = 0,2
[H+] = a . M = 1. 0,2 = 0,2 = 2.10^-1
pH = - log H+ = - log 2.10^-1 = 1 - log 2
c. HCl 0,01 M
jawab : [H+] = a. M = 1 . 10^-2 = 10^-2
pH = - log H+ = - log 10^-2 = 2
d. HNO3 0,006 M
jawab : [H+] = a . M = 1. 6.10^-3 = 6.10^-3
pH = - log H+ = - log 6.10^-3 = 3 - log 6
e. 9,8 gram H2SO4 dalam 100 ml air (Ar H=1, Ar S=32, Ar O=16)
jawab : mol H2SO4 = massa/Mr = 9,8/98 = 0,1
M = mol/volume = 0,1/0,1 = 1
[H+]= a .M = 2 . 1 = 2
pH = - log H+ = - log 2
f. 10 gram HCl (Ar H=1. Ar Cl = 35,5) dalam 200 ml air
jawab : mol = massa/Mr = 10 / 36,5 = 0,27
M = mol/volume = 0,27/0,2 = 1,35
[H+] = a . M = 1 . 1,35 = 1,35
pH = - log H+ = - log 1,35
g. Tentukan massa HCl yang memiliki pH = 3 - log 4 jika volumenya 500 ml
jawab : pH = 3 - log 4
[H+] = 4 . 10^-3
mol = M . volume = 4.10^-3 . 5.10^-1 = 2.10^-3
massa = mol . Mr = 2.10^-3 . 36,5 = 73.10^-3 = 0,073 gram