Amonium Hidroksida
Mr =Ar N + 5Ar H + Ar O =14+5+16 =35
jumlah mol(n) =m/Mr =10/35 =0,286 mol
Konsentrasi(M) =n/V =0,286 mol/3 L =0,095 M

derajat ionisasi(alpha) =akar (Kb/Mb)
=akar(4,9x10^-5/0,095)
=akar(0,0005)
=0,0223

[OH-]=akar(KbxMb)
=akar(4,9x10^-5 x 0,095)
=akar(0,00000466)
=0,00216 M

Jawabanmu

2014-04-14T17:50:39+07:00
Mr NH4OH = (AR N) + (4. AR H) + (AR O)+(AR H) = 14 + 4.1 + 16 + 1 = 35
2014-04-14T17:52:27+07:00
AR= N= 14    H= 1    0=16
MR= 1XN + 4XH + 1XO + 1XH
     =  14 + 4 + 16 + 1
     =  35
semoga membantu :)