计算C（C7H4O3Na2）＝0.050摩尔每升的水杨酸钠溶液的PH值以及溶液中的各物质的浓度已知水杨酸的Ka1 Ka2_百度作业帮
计算C（C7H4O3Na2）＝0.050摩尔每升的水杨酸钠溶液的PH值以及溶液中的各物质的浓度已知水杨酸的Ka1 Ka2
计算C（C7H4O3Na2）＝0.050摩尔每升的水杨酸钠溶液的PH值以及溶液中的各物质的浓度已知水杨酸的Ka1 Ka2
好难呀我爸爸想了好久C7H4O32- + H2O C7H4O3H- + OH- Ka2=1.82 x 10-14 C7H4O3H- + H2O C7H4O3H2 + OH- Ka1=1.07 x 10-3 Since Ka1 and Ka2 have a difference larger than 10^10 times,in the hydrolysis,first step contributes to [OH-] most,and hence to the pH.So to calculate pH we only need to consider the first step of reaction.That is:C7H4O32- + H2O C7H4O3H- + OH- Kb1=Kw/Ka2=0.549 = ([C7H4O3H-][OH-])/[C7H4O32-]=x^2/(0.050-x) x=[OH-]=[C7H4O3H-]=0.046 M ==>pOH=1.34 ==>pH=14.00-1.34=12.66 Now if you like,you can calculate [C7H4O3H2],which is the product of second step of hydrolysis,and will be extremely small.Here is the calculation Kb2=Kw/Ka1=9.35 x 10^-12 = (([C7H4O3H2][OH-])/[C7H4O3H-]=x'(0.046+x')/(0.046-x') x'=[C7H4O3H2]=9.35 x 10^-12 M So total [OH-] =x +x' =0.046 + 9.35 x 10^-12 =0.046 M and [C7H4O3^2-]=0.050-x=0.050-0.046=0.004 M so you got them all.Cheers BTW,if the question also gives pH.then you can calculate everything accurately.LS's answer is only suitable for Kb1> 10^-5
C7H4O32- + H2O
C7H4O3H- + OH-Ka2=1.82 x 10-14C7H4O3H- + H2O
C7H4O3H2 + OH-Ka1=1.07 x 10-3Since Ka1 and Ka2 have a difference larger than 10^10 times, in the hydrolysi...
[OH-]=(0.05Kw/Ka2)^1/2 ; (开根号)一摩尔每升醋酸及醋酸钠配制PH值为四的缓冲溶液十毫升_百度作业帮
一摩尔每升醋酸及醋酸钠配制PH值为四的缓冲溶液十毫升
一摩尔每升醋酸及醋酸钠配制PH值为四的缓冲溶液十毫升
近似公式：pH=pKa+lg（cAc/cHAc）醋酸pKa=4.74 代入则有lg（cAc/cHAc）=-0.74 由此处浓度比等于相同浓度所取体积比,故VAc：VHAc=10^(-0.74)=0.182则所用醋酸约8.46mL,醋酸钠约为1.54mL
近似计算8.2ml醋酸1.8ml 醋酸纳……假设x ml 醋酸纳根据电离方程……H +浓度为10∧-4醋酸浓度近似为（10-x ）/10 醋酸根近似为x /10……利用醋酸离子积为1.8×10∧-5……解得x 约为1.8欲从0.0020摩尔每升pb(No3)2溶液中产生pb(oH)2沉淀,溶液的PH至少为多少_百度作业帮
欲从0.0020摩尔每升pb(No3)2溶液中产生pb(oH)2沉淀,溶液的PH至少为多少
欲从0.0020摩尔每升pb(No3)2溶液中产生pb(oH)2沉淀,溶液的PH至少为多少
好多符号不好打就截了个图,0.1mol/L的醋酸和 0.1mol/L 的醋酸钠等体积混合稀释10被的溶液PH值为多少?怎么计算?_百度作业帮
0.1mol/L的醋酸和 0.1mol/L 的醋酸钠等体积混合稀释10被的溶液PH值为多少?怎么计算?
0.1mol/L的醋酸和 0.1mol/L 的醋酸钠等体积混合稀释10被的溶液PH值为多少?怎么计算?
当然有办法计算：根据醋酸电离常数K=1.8×10-5HAcH+ ＋Ac-K=[H+][Ac-]/[HAc]忽略电离及水解产生的[Ac-]、[HAc](因为相比少得可忽略),[HAc]=[Ac-]=0.01mol/L[H+]=1.8×10-5,pH=4.75
这个真的没法计算
为什么不能算？
因为你给的条件不全啊摩尔 制备 溶液ph_百度作业帮
摩尔 制备 溶液ph
摩尔 制备 溶液ph}