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 crys
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.
化学人气:547 ℃时间:2020-05-08 01:42:37
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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)-l...我想问一下是不是10的负0.3我怕我理解错了、、是^就是幂的意思,世界通用但是你写的是0.3 不是负0.3我怕我计算错了 我感觉是负0.3哦哦 是负的哈,我少写个符号我算的是错误的你算-0.3是对的。10^-0.3=0.5012后面的也是0.75X0.5012mol 不是1.9953。0.0犯个低级错误哈哈
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