ke dalam 250 ml larutan H2SO4 yg Ph nya 2- log 4 dialirkan 672 ml gas NH3 (STP). Tentukan Ph larutan? Kb = 1,8 X 10^-5

1) Cari n H2SO4

PH = 2-log 4
[H⁺] = 4*10⁻²
M = 2*10⁻²
n = MV = 2*10⁻² * 250 = 5 mmol

2) Cari n NH3
n = V/22,4 = 672/22,4 = 30 mmol

3) buat mrs

H2SO4 + 2NH4OH -> NH42SO4 + 2H2O
m : 5 mmol 30 mmol – –
r = 5 mmol 10 mmol 5 mmol 10 mmol
s = – 20 mmol 5 mmol 10 mmol

karena basa lemah dan garamnya yang tersisa maka gunakan rumus buffer basa

4) masukkan data ke rumus
[OH⁻] = Kb[NH4OH]/[NH42SO4]
= 1,8*10⁻⁵ [20]/[5]
= 72 x 10⁻⁶

5) cari POH
POH = -log[OH⁻]
POH = 6-log72
POH = 4,14

6) cari PH
PH = 14-POH
PH = 14-4,14 = 9,85 / 8+log 72