2605 pH = 9–> pOH = 14 – 9 = 5–> OH- = 10^-52605garam = 10^-12605Kb = 2 \xd7 10^-52605 gunakan rumus larutan penyangga basaOH- = Kb \xd7 NH3 / garamNH3 = OH- \xd7 garam / Kb= 10^-5 \xd7 10^-1 / 2 \xd7 10^-5NH3 = 5 \xd7 10^-2 <------
2605 pH = 9–> pOH = 14 – 9 = 5–> OH- = 10^-52605garam = 10^-12605Kb = 2 \xd7 10^-52605 gunakan rumus larutan penyangga basaOH- = Kb \xd7 NH3 / garamNH3 = OH- \xd7 garam / Kb= 10^-5 \xd7 10^-1 / 2 \xd7 10^-5NH3 = 5 \xd7 10^-2 <------