这个问题很简单,可是查阅了好久才找到靠谱方法!来自下面这本书:<br>Essential Laboratory Mathematics: Concepts and Applications for the Chemical<br>example:文献中会提出 配制<br>首先0.2M(mol/L)是指 醋酸钠加上醋酸的mol/g之和<br>所以设c(HAc)为Xmol/L,则c(NaAc)为(0.2-X)mol/L.(欲配得溶液中)<br>然后,这两个浓度不论在初始配制和配得溶液中须满足下面方程:<br>例中要求pH=5.2,Ka是HAC乙酸的解离常数,查表。<br>此处pKa=-logKa,常数,为4.75<br>则5.2=4.75+log((0.2-X)/X)<br>所以 c(NaAC)/c(HAc)=2.82<br>X+2.82X=0.2<br>解X=0.052 mol/L,所以c(HAc)为0.052mol/L, c(NaAC)为0.15mol/L.<br>需要mHAc =cM=0.052*60=3.12g/L,通常为液体,密度1.05g/mL. 配1L需要体积2.97mL.<br>mNaAc=cM=0.15*82=12.3g/L<br>若是三水醋酸钠,M=136=20.4g/L<br>所以要配制1L缓冲液,需要20.4g 三水醋酸钠,2.97mL无水乙醇。定容至1000mL