配置100公斤ph为3.2酸度的酸水需要37%的盐酸多少ml

2025-04-06 19:09:29
推荐回答(1个)
回答1:

浓盐酸质量浓度是37%,密度1.19g/mL
pH=3.2则C(H+)=10^(-3.2)=6.31X10^(-4)mol/L
m(HCl)=n(HCl)*M(HCl)=n(H+)*M(HCl)=C(H+)*V(H+)*M(HCl)
=6.31X10^(-4)*[100X10^3/1.0*10^(-3)]*36.5=
=2.3
V(HCl)=m(HCl)/p=2.3/1.19=1.93mL