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m 20 50 - -
r 20 20 20 20
s - 30 20 20
Asam lemah dan garam yang tersisa ->> Buffer asam
[H⁺] = Ka [Al]/[g]v
[H⁺] = 10⁻⁵ [30]/20x1
[H⁺] = 1,5 x 10⁻⁵
PH = 5- log 1,5
2) HNO3 + NH4OH ->> NH4NO3 + H2O
m 8 32 - -
r 8 8 8 8
s - 16 8 8
Basa lemah dan garam yang tersisa ->> Buffer basa
[OH⁻] = Kb [Bl]/[g] v
[OH⁻] = 10⁻⁵ [16]/8 x 1
[OH⁻] = 2 x 10⁻⁵
POH = 5-log 2
PH = 9 + log 2
3) PH akhir = 2 x PH awal = 2 x3 = 6
->> [H⁺] = 10⁻⁶
[H⁺] = Ka [Al]/[g] x V
10⁻⁶ = 10⁻⁵ [10²]/[g]xV
10⁻⁶ = 10⁻³/[g]
[g] = 10⁻³/10⁻⁶ = 10³ mmol
Mr natrium asetat = 72
massa = nMr = 72 x 10³ mg atau 72 gram