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Rumus pH Campuran :

[Asam] = (V1 × M1) + (V2 × M2) + (V3 × M3) / V1 + V2 + V3

[Asam] = (50 × 0,2 M) + (50 × 0,1 M) + (75 × 0,1 M) / 50 + 50 + 75

[Asam] = (10,0) + (5) + (7,5) / 175 mL = 0,12 mol

Rumus pH = -Log[H+]

pH = -Log[ 12 × 10^-2 ]

pH = ( 2-Log12 )

pH = ( 0,9 )

pOH = 14 - (-Log [OH-] )

pOH = 14 - (2-Log 12 )

pOH = 14 - ( 0,9 )

pOH = 13,1

pertama2 kita cari dahulu jumlah mol nya :

Mol HCl = 0,1 M × 50 mL = 5 mmol

Mol HBr = 0,2 M × 50 mL = 10 mmol

Mol NaOH = 0,1 M × 75 mL = 7,5 mmol

HCl + HBr + NaOH → H2O + HBr + NaCl

Mula : 5 : 10 : 7,5 → ... : ... : ...

Reaksi : 5 : 5 : 5 → 5 : 5 : 5

Sisa : 5 + 2,5 = 7,5 mmol

[ NaOH ] = 7,5 mmol / 175 mL = 0,04 molar

pOH =. 14 - (-Log [ 4 × 10^-2 ])

pOH = 12,61

pH = (-Log [ 4 × 10^-2 ])

pH = 2 -Log 4

pH = 1,4

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