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
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,not for here since Kb1=0.549 >> 10^-5

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 hydrolysi...

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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, not for here since Kb1=0.549 >> 10^-5

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[OH-]=(0.05Kw/Ka2)^1/2 ; (开根号)