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

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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.
 
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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.