Explain how to make 5L of 0.15M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystalline sodium acetate.

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Explain how to make 5L of 0.15M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystalline sodium acetate.
Explain how to make 5L of 0.15M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystalline sodium acetate.

Explain how to make 5L of 0.15M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystalline sodium acetate.
原题翻译.
解释如何使用1.0mol/L的醋酸和醋酸钠晶体（固体）配置5L 0.15mol/L的醋酸-醋酸钠缓冲溶液.
醋酸PKa=4.7（这个是常识不解释）
c(H3O+)=Ka(HA)*c(HA)/c(A-)
-lg(c(H3O+))=-lgKa(HA)-lg(c(HA)/c(A-))
PH=pKa(HA)-lg(c(HA)/c(A-))或PH=pKa(HA)+lg(c(A-)/c(HA))
欲配置PH=5.0的缓冲溶液则
5.0=4.7-lg(c(HA)/c(A-)),c(HA)/c(A-)=10^0.3=1.9953 为了配置5L这种缓冲液,且要求0.15mol的醋酸醋酸钠溶液.需要醋酸(HA)0.75mol 已知有的是1.0mol/L的醋酸则加入0.75L
因为 c(HA)/c(A-)=10^0.3=1.9953 所以还需要加入0.75X1.9953mol的醋酸钠固体.然后用水定容到5L即可.得到所需要的5L 0.15mol/L的醋酸-醋酸钠缓冲溶液.
楼上的这个确实不是高中化学能解答的,这个是大学化学哈

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