a. 50 ml CH3COOH 0,1 M + 25 ml KOH 0,1 M
b. 25 ml CH3COOH 0,1 M + 50 ml KOH 0,1 M
c. 50 ml CH3COOH 0,1 M + 50 ml KOH 0,2 M
d. 50 ml CH3COOH 0,1 M + 25 ml KOH 0,1 M
e. 25 ml CH3COOH 0,1 M + 50 ml KOH 0,1 M
Larutan buffer di bawah ini yang mempunyai pH = pKa adalah
PH = pKa + log [garam]/ [asam]
pH = pKa , log [garam] / [asam] harus 0 atau [garam] = [asam]
a mmol asam = 50 x 0,1 = 5 mmol
mmol garam = 5 – (25 x 0,1) = 5 – 2,5 = 2,5 mmol
sisa mmol asam = 5 – 2,5 = 2,5 mmol
[garam]/[asam] = 1 => pH = pKa
b. mmol asam = 25 x 0,1 = 2,5 mmol
mmol garam = 2,5 -(50×0,1) = 2,5 – 5 = -2,5 ( ada sisa basa) => salah
c. mmol asam = 50 x 0,1 = 5 mmol
mmol garam = 5 – (50×0,2) = 5 – 10 = -5 (ada sisa basa) => salah
d. mmol asam = 50 x 0,1 = 5 mmol
mmol garam = 5 – (25×0,1) = 5 – 2,5 = 2,5
sisa mmol asam = 5 – 2,5 = 2,5 mmol
[garam]/[asam] = 1 => pH = pKa
e. mmol asam = 25 x 0,1 = 2,5 mmol
mmol garam = 2,5 – (50 x0,1) = 2,5 – 5 = -2,5 (ada sisa basa) => salah
jawabannya a & d karena jumlah ml dan konsentrasinya sama