Odpowiedź:
n-liczba moli
V-objętość
k-kwasu
z-zasady
Bufor kwasowy, równanie Hendersona-Hasselbacha:
pH=pka-log(Ck/Cz) gdzie C=n/V zatem V=const
pH=pka-log(nk/nz)
Ka=1,8*10^(-5) zatem pHa=-logka=-log(1,8*10^(-5)=4,745
1.
n1=0,1*18,8=1,88mmol
n2=0,1*1,2=0,12mmol
pH=4,745-log(1,88/0,12)=3,55
2.
n1=0,1*16,4=1,64mmol
n2=0,1*3,6=0,36mmol
pH=4,745-log(1,64/0,36)=4,09
3.
n1=0,1*11,2=1,12mmol
n2=0,1*8,8=0,88mmol
pH=4,745-log(1,12/0,88)=4,64
4.
n1=0,1*6=0,6mmol
n2=0,1*14=1,4mmol
pH=4,745-log(0,6/1,4)=5,11
5.
n1=0,1*4=0,4mmol
n2=0,1*16=1,6mmol
pH=4,745-log(0,4/1,6)=5,35
6.
n1=0,1*2,4=0,24mmol
n2=0,1*17,6=1,76mmol
pH=4,745-log(0,24/1,76)=5,61
7.
n1=0,1*2=0,2mmol
n2=0,1*18=1,8mmol
pH=4,745-log(0,2/1,8)=4,7
