Salt of weak acid and base

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Salt of weak acid and base

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  • #1
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Summary:

Because of character limit I have exposed my question in the attached pdf
Because of character limit I have exposed my question in the attached pdf
 

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  • #2
symbolipoint
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I wish I could view the file WITHOUT downloading it. I will NOT download the file.
 
  • #3
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Let me explain: at the beginning I thought I had to insert the post in the "summary" box and it gave me a limit of 250 characters, so I wrote the post in the form of a PDF attachment. I'll rewrite it. Please help me if you can.
In general, when dealing with a weak acid salt and weak base, it is assumed that the three reactions
(hydrolysis of BH+,of A-and self water autolysis) depend on each other and there for only one degree
of hydrolysis is used. In reality if the acidity constant of the base is different from that of the basicity of
the acid the degree of ionization of BH+ is different from that of A-. Assuming that Cs is the initial
concentration of the BHA salt then(BH+)i=(A-)i=Cs. At equilibrium [BH+]=Cs-x or for x=Csα1 then
[BH+]=Cs(1-α1) and [ B]=[H3O+]=x; At equilibrium [A-]=Cs-y or for y=Csα2 then [A-]=Cs(1-α2)
and [HA]=[OH-]=y; If x is different from y how is the pH calculated?Note that Ka. of the base=([ B][H3O+])/[BH+] and Kb of the acid=([HA[OH-])/[A-] where Ka of the acid=Kw/Kb of the acid.Please help me!!
 
  • #5
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Hello Borek, I read the link and I was inspired but I made some changes because in my case [BH +] = Cs- and [A -] = Cs- [HA]. From the charge balance [A -] + [OH-] = [BH +] + [H3O +]. The Ka of BH + is ([H3O +]) / [BH +] = ( [H3O +]) / (Cs- ) and for Cs >> we have that Ka of BH + is equal ad ([H3O +])/ Cs; similarly the Ka of HA is ([A -] [H3O +]) / [HA] = ((Cs- [HA]) [H3O +]) / [HA]; and for Cs >> [HA] we have that Ka of HA is equal to (Cs [H3O +]) / [HA]. So = Ka di BH+ * Cs / [H3O +]; [HA] = (Cs * [H3O +]) / Ka of HA. Always from the charge balance (Cs - [HA]) + Kw / [H3O +] = (Cs - ) + H3O +. Cs - (Cs * [H3O +] / Ka of HA) + Kw / [H3O +] = Cs - (Ka of BH + * Cs / [H3O +]. From which - (Cs * [H3O +] / Ka of HA) + Kw / [H3O +] = - (Ka of BH + * Cs / [H3O +]. Therefore, rearranging (Ka of BH + * Cs / [H3O +] + Kw / [H3O +] = (Cs * [H3O +] / Ka of HA) + [ H3O +]. Applying the logarithms at the end results that 4pH = (pka of BH +) + (pka of HA) + pkw I honestly do not believe that this is the result.
 
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  • #6
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I don't know why it never appears in the post.
 
  • #7
Borek
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[ B] marks the beginning of the bolded text, add a space after the opening square bracket to to avoid that.

Please try to use LaTeX to format your equations (there is a link just below the edit field that explains how to do so):

[tex]K_a = \frac {[H^+][A^-]}{[HA]}[/tex]

I am afraid at the moment your post is unreadable so quite difficult to follow.
 
  • #8
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Hello, today taking inspiration from you, I went to review the demonstration of the polyprotic acids and I finally succeeded in extracting that [H3O +] = {(Kw + Ka of the base * [BH +]) / (1+ [A -] / Ka acid)} ^ 1/2. If [BH +] ~ Cs and [A-] ~ Cs then [BH +] = [A -] = Cs. Then [H3O +] = {(Kw + Ka of the base *Cs) / (1+ Cs / Ka of the acid)} ^ 1/2. For Ka of the base * Cs >> Kw and Cs / Ka of the acid >> 1 then [H3O +] = {(Ka of the base * Cs) * (Ka of the acid/Cs)} ^ 1/2. So [H3O +] = (Ka of the base * Ka of the acid/Cs)^ 1/2.
 

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