一道关于离子平衡的化学题,

令一元酸为HA
其钠盐的物质的量=36.12mL*0.10mol/L=3.612mmol
再加入盐酸后:
n(HA)=18.06mL*0.10mol/L=1.806mmol
剩余钠盐:n(A-)=3.612mmol-1.806mmol=1.806mmol
显然形成缓冲溶液
HA H+ +A-
Ka=[H+]*[A-] /[HA]
=约=[H+]*c(A-)/cHA)
=[H+]*n(A-)/n(HA)
=[H+]*1.806mmol/1.806mmol
=[H+]
=10^-4.92=1.2*10^-5
==
也可用:
pH=pKa +lgc(A-)/cHA)
=pKa +lgn(A-)/n(HA)
=pKa
因此pKa=pH=4.92
Ka=1.2*10^-5