24.06.2010, 21:59
0,4g NaOH werden in 100ml Ammoniumsulfat (c=0,1, Kb=1,1x10^-5) gelöst. Welchen pH-Wert hat die resultierende Lösung
n(NaOH)= m/M =0,4/40 =0,01mol
n[(NH4)2SO4]=c.V =0,1.0,1=0,01mol
pKB=4,96 (-log1,1x10^-5)
pKS=9,04 (14-pKB)
pH=pKs+log [Base]/[Säure]
pH=9,04
n(NaOH)= m/M =0,4/40 =0,01mol
n[(NH4)2SO4]=c.V =0,1.0,1=0,01mol
pKB=4,96 (-log1,1x10^-5)
pKS=9,04 (14-pKB)
pH=pKs+log [Base]/[Säure]
pH=9,04