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pOH: -log OH- : 5-log2
pH : 14-pOH : 14-5-log2 : 9+log2
NH₃ = 0,16 x 50 = 8 mmol
NH₃ + HCl --> NH₄Cl
mula-mula 8 8 -
bereaksi 8 8 8 ₋
setelah reaksi - - 8 mmol
n NH₄Cl = 8/100 = 0,08
[OH⁻] = √10⁻¹⁴ . 8 x 10⁻²
2 x 10⁻⁵
= √4 x 10 ⁻¹¹
= 2 x 10⁻⁵·⁵
POH= -log[OH⁻]
= -log 2 x 10⁻⁵·⁵
= 5,5 - log 2
pH = 14-(5,5-log2)
= 8,5 + log 2
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