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Chapter 9

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Chapter 9

Systematic Treatment of Equilibrium

- Basic concept of electroneutrality
- Sum of the positive charges in solution equals the sum of the negative charges in solution.

- Ions in solution.
- What species will ionize?
- Sodium Chloride (NaCl) will dissolve into solution to give us Na+ and Cl- in equal amounts.
- Glucose will dissolve in solution but not generate ions.

- H3PO4
- It is time to recall some chemistry.
- We know that PO43- will result but what else.
- H+ to be sure.
- Phosphate goes thru a series of equilibrium steps to give a range of species
- H2PO4- , HPO42- , PO43- that have charge.

- Note: Species with alkali metal ions combined will not really exist. (There will be no NaHPO4- for example)

- Since it is a water solution and we know that water can dissociate then we would have OH- also.
- So we would have H+ for cations.
- And OH-, H2PO4- , HPO42- , PO43- for anions.

- Scations = Sanions
- Ions of charge greater than one must be accounted for. So for example if we have a calcium ion formed then each calcium ion has a double charge we must multiply the concentration of such an ion by this charge.

- So for phosphoric acid we have the following charge balance.
- [H+] = [H2PO4-] + 2[HPO4=]+3[PO43-]+[OH-]
- The H3PO4 that remains in solution has no charge and need not be accounted for in the charge balance.

- What if we put trisodium phosphate into solution. Na3PO4
- This is the salt of a strong base and the sodium can be assumed to completely dissociate. (We know what ion pairs are and we will elect to ignore them unless directed otherwise.)

- Well we will have Na+ and PO43-.
- But!!!
- Phosphateion is a fairly strong base. That is it would love the rob a proton from somewhere and become HPO42-
- This ion in turn is a relatively strong base too and will steal another proton (H2PO4-)
- This new ion is also capable of the next step giving us H3PO4.

- So for this solution we will possibly have.
- PO43- , HPO42- , H2PO4- ,H3PO4 and Na+
- And the ions from water. H+ and OH-
- So our charge balance is
- [H+] + [Na+] = [H2PO4-] + 2[HPO4=]+3[PO43-]+[OH-]

- If we were to put K2PO4 into solution then what would we expect to see for relative amount of each of these species.

- I do not want to see species like NaPO42- cropping up in charge balances. They do not exist! This is usually a pit fall for you all in preparing your charge balances.

- All this tells us that what be put into our solution is in there someplace. This is our statement of conservation of mass.
- So if we put Phosphoric acid into solution it will be in there as one of the phosphate species. Phosphoric acid is an ingredient of diet pepsi.
- We know that it undergoes these acid base interactions so our Mass balance accounts for that.

- CH3PO4 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
- This CH3PO4 is often referred to as the Formal Concentration. So if we put 1.5 mmoles of phosphoric acid into 0.50 liters of water then this formal concentration would be 3.0 mM.
- We must account for each species, charged or not, and since we are looking at concentration and not charge we should not multiply by the charge in this case.

- If we do not know the concentration we can still use the mass balance.
- When we put this salt into water we know we get three sodium for each phosphate. To set this up into an equation we could write.
- [Na+] = 3 * phosphate concentration
- Have I written this backward?????

- So we would have a mass balance of
- [Na+]= 3{[H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]}

- Write all pertinent reactions (don’t forget water)
- Write the charge balance
- Write the mass balances ( there might be more than one )
- Count the equations and unknowns. You will need to have and equal number here.
- Now just solve!

- Let us look at the trisodium phosphate.
- Unknowns
- [Na+] [H3PO4] [H2PO4-] [HPO42-] [PO43-] [H+] [OH-]

- Seven Unknown so we need seven equations.
- Mass balance on sodium ion
- Mass balance on phosphate ions
- Kw
- Charge balance
- Three acid equilibria

- Full solution will be a seventh order polynomial.

- That was just a simple system – these equilibrium systems can become very complex and solution depends on our knowing all the equilibria involved.

- Two equilibria involved
- Ksp
- Kb for F- once it is dissolved
- Setting conditions can help solve this problem