Question

Sebanyak 100 ml NH4Cl 0,1 mempunyai pH=9, maka berapakah massa amonia? (kb=2 x 10 pangkat -5, Ar N=14, H=1)

Answer 1

Diket:
*n NH4Cl = n NH4+ = 100 mL x 0,1 M = 10 mmol = 0,01 mol
*pH = 9 --> [OH-] = 10^-5 
*Kb = 2 x 10^-5
*Mr NH3 = 17

Ada NH4Cl (garam) dan NH3 (basa lemah), maka memakai rumus buffer:
[OH-] =  Kb x mol NH3/mol NH4+
10^-5 = 2 x 10^-5 x mol NH3 / 0,01 mol
0,01 mol = 2 x mol NH3
mol NH3 = 0.005 gram/Mr

Massa NH3 = n x Mr
= 0,005 x 17
= 0.085 gram

Answer 2

----> 
mol=0,1 L.0,1 M
     = mol
pH=9 ---> pOH=5