Dissociation type problem

  • #1
I think that is what you call it. I need to find the pH of a solution containing 50 grams of Na3PO4.

50 g ---- > 0.305 M

We know that Kw=Ka*Kb

So,

Na3PO4 + H20 --> HNa3PO4 + OH (I think)

Kb = x^2/(.305-x)

I cant find the value of Kb anywhere in my book, internet. If I had it I could calculate Ka and get my pH. Otherwise Im stumped.
 

Answers and Replies

  • #2
dextercioby
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Do you mind if i'm asking you to write the dissociation relation correctly...?

Daniel.
 
  • #3
MMM? Doent 10^-14 = Kw = Ka*Kb

and Na3PO4 + H20 --> HNa3PO4 + OH
(Base) (Acid) (Con A) (Con B)

Initial: .305M -- 0 0
Change -x -- +x +x
Final: .305 -x -- x x

Kb = products/reactants = [x][x] / [.305 - x]

If I knew what the value of Kb was for Na3PO4, I could get x and calculate the pOH and ofcourse get the pH.
 
  • #4
dextercioby
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The chemical (well,ionic) reaction is incorrectly written...Think it viceversa:the base +acid--->Na phosphate+water.

Daniel.
 
  • #5
The problem doesnt mention any specific acid and all the preceeding problems use water in the reactants. Anyhow, cant water be considered an acid (thats browsy rule or something)?

I thought the Base would pick up an H+ and since the H20 lost an H, it would simply become hydroxide
 
  • #6
dextercioby
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I saw it like
[tex]Na_{3}PO_{4}+3H_{2}O\rightleftharpoons H_{3}PO_{4}+3NaOH [/tex]

Anyway,pay attention with the solving and about the data that u're missing,i don't know,maybe someone else would guide u better...

Daniel.
 
  • #7
Gokul43201
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The hydrolysis may not be complete, and so you might also have Na2HPO4 and NaH2PO4. From thie equilibrium constants you can tell which one (if any) is dominant.
 
  • #8
GCT
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In cases such as these you observe the corresponding cationic and anionic components of the salt, the pH will differ according to whether the conjugates of such are strong acid/bases. In this case the base will contribute to the pH. Usually for these types of problems, we can neglect matters regarding the formation H2PO4 since the conjugate base of it is a fairly weak base.

what's this?
50 g ---- > 0.305 M
Why don't you try stating the problem word by word.
 
  • #9
#80.

Trisodium phosphate (Na3PO4) is available in hardware stores as TSP and used as a cleaning agent. The label on a box of TSP warms that the substance is very basic. What is the pH of a solution containing 50.0 g of TSP per liter?
 
  • #10
Gokul43201
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Do you see why it's important to not leave out the "per liter" ?
 
  • #11
GCT
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I suppose that they want us to assume that it is a strong base

-The net equation will consist of aqueous [tex]PO_4^-3[/tex] reacting with a hydronium cation [tex]H_3O^+[/tex] to produce [tex]HPO_4^-2[/tex] and [tex]H_2O[/tex] Write and balance out this equation.

-You can than deduce the final concentration of [tex]OH^-[/tex] from finding the initial molarity of the [tex]PO_4^-3[/tex], there is a stochiometric one to one equivalence of the compound to this anion, so find the initial molarity of the compound and this will be the concentration of the anion.

-Use the balanced equation to find the stoichiometric ratio of the anion to the hydroxide.

-Use this to find the [tex]p_{OH}[/tex] then deduce [tex]p_H[/tex]
 
  • #12
Borek
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GeneralChemTutor said:
The net equation will consist of aqueous [tex]PO_4^-3[/tex] reacting with a hydronium cation [tex]H_3O^+[/tex] to produce [tex]HPO_4^-2[/tex] and [tex]H_2O[/tex].
And why not [tex]PO_4^3^- + H_2O \rightarrow HPO_4^2^- + OH^-[/tex]?

Due to a very small third dissociation constant you may assume that the equilibrium is shifted 100% to the right - the you may proceed as you have proposed, but this way you don't have to introduce [tex]H_3O^+[/tex] ions.


Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis
 
  • #13
GCT
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The base will react with the strongest acid.
 
  • #14
GCT
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That is the base increases the pH level by first reacting with the strongest acid , the [itex][H^+][/itex] level decreases as a result. You can guess what happens next.
 

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