双组分溶液体系中[H^+]计算与讨论  

[H^+]Calculation and Discussion of Solutions Having Two Components

作　　者：赵长春[1] 卢菊生[1] 王颇[1] 刘晓君[1] 

机构地区：[1]江苏师范大学化学化工学院,江苏徐州221116

出　　处：《广州化工》2016年第6期127-129,共3页GuangZhou Chemical Industry

基　　金：江苏高校品牌专业建设工程项目资助

摘　　要：对双组分体系中[H^+]的计算做了理论推导,并对简化条件进行讨论。对于强酸~弱酸体系,H^+=(c_(HX)-K_a)+((c_(HX)-K_a)~2+4K_a(c_(HX)+c_(HA)))^(1/2)/2,当c_(HX)/c_(HA)+K_a≥20K_a/c_(HX)+K_a时,可用最简式[H^+]=c_(HX)。对于缓冲溶液体系,从物料平衡和质子等衡角度出发,推导出计算[H^+]的精确式,[H^+]=K_ac_a-[H^+]+[OH^-]/c_b+[H^+]-[OH^-],使得公式中c_a和c_b的概念更好地理解。对于弱酸~弱酸体系,体系中HA为较强的弱酸且两者的分析浓度接近时,此时K_(HA)/K_(HB)<10且满足c_(HA)/K_(HA)≥0.5或者10<K_(HA)/K_(HB)<100成立且满足c_(HA)/K_(HA)≥4,可以采用近似式[H^+]~2=[H^+]~2_(HA)+[H^+]~2_(HB)+K_W。The [H＋] calculations in different solutions having two components and the approximate discriminant equations were discussed under consideration. For the mixture of strong acid and weak acid,H＋=（c_（HX）-K_a）＋（（c_（HX）-K_a）2＋4K_a（c_（HX）＋c_（HA）））（1/2）/2,when it met the condition of c_（HX）/c_（HA）＋K_a≥20K_a/c_（HX）＋K_a,[H＋]= c_（HX）. For the buffer solutions,in order to understand the concept of c_aand c_bin the equation,we got the [H＋]=K_ac_a-[H＋]＋[OH-]/c_b＋[H＋]-[OH-]from the MBE and PBE. For the mixture of weak acids,if the analytical concentration was closed to each other,K_（HA）/ K_（HB）10and c_（HA）/ ＋K_（HA）≥0. 5 or 10K_（HA）/ K_（HB） 100 and c_（HA）/ K_（HA）≥4,[H＋]2=[H＋]2_（HA）＋[H]2_（HB）＋K_Wcan be used as approximate discriminant equation.

关 键 词：质子等衡式 强酸弱酸混合溶液 缓冲溶液 弱酸弱酸混合溶液 

分 类 号：O655.2[理学—分析化学]