0.05mol/L醋酸缓冲液(ph=5)如何配置

题目

0.05mol/L醋酸缓冲液(ph=5)如何配置
要配0.05M的醋酸缓冲液150ML,PH=5.0,PKA=4.77,用0.5M的醋酸和6N 的氢氧化钠配

答案

[H+]*[Ac-]/[HAc] =10^-4.77
[[Ac-]/[HAc] =10^-4.77/10^-5 =10^0.23
[[Ac-]= 10^0.23 *0.05
[HAc] =0.05M
6V(NaOH)=10^0.23*0.5*150
V(NaOH)=12.5*10^0.23ml
{[10^0.23 *0.05+0.05M}*150=0.5V(HAc)
V(HAc)={[10^0.23 *0.05+0.05M}*300ml