Calculating pH of Na3PO4 (45.0g/L): Find the Basicity

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Discussion Overview

The discussion revolves around calculating the pH of a sodium phosphate solution (Na3PO4) at a concentration of 45.0 g/L. Participants explore the basicity of the solution, the dissociation of the compound in water, and the implications of treating it as a strong versus weak base.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Joanna calculates the number of moles of Na3PO4 and assumes it completely dissociates in water, leading to a calculated pH of 13.431.
  • Some participants argue that PO4-3 is not a strong base and suggest that the pH would be slightly lower, indicating it should be treated as a weak base.
  • Another participant emphasizes the need to incorporate the equilibrium constant in the calculations.
  • There is a suggestion that without the equilibrium constant, participants must find alternative methods to approach the problem without using an ICE table.
  • One participant expresses skepticism about the assumption of complete dissociation and acknowledges the potential for error in the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether Na3PO4 should be treated as a strong or weak base, leading to differing views on the calculated pH. The discussion remains unresolved regarding the correct approach to determining the pH.

Contextual Notes

Limitations include the absence of the equilibrium constant and the lack of clarity on whether to treat Na3PO4 as a strong or weak base, which affects the pH calculation.

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We have 45.0g of Na3PO4 per liter. We know it is very basic but need the pH.
I don't have the answers to all exercises, so if you could check my work, that'd be great.

So, we have g, but moles would be more useful.
So 45g x (1 mol/163.937) = 2.7 x 10^-1 mol.

We know that strong bases completely dissociate in water.
Therefore, [Na3PO4]=[OH-].
So, since pOH = - log of all that = 0.569
Therefore: pH = 14 - 0.569 = 13.431 which is indeed pretty basic.

Any mistake somewhere?

Thank you,

Joanna.
 
Last edited:
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You need to incorporate the equilibrium constant toward solving this problem.
 
We do not have the equilibrium constant, and if they don't give us and don't tell us to refer to a table, it means we have to find a way around it (without an ICE table)...
 
No way. You will either use dissociation constant or it'll be pure guesswork.
 
That's exactly why I assumed that, since they said that it was very basic, it completely dissociated in H2O, and it's why I went on with calculations without an ICE table.

Well, I'll just assume it's right and see what the correction is when we get it.
 

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