[Shadow]

What indicator you could use for the titration of NaH₂PO₄ (c=0.1mol/dm³) with NaOH (c=0.1mol/dm³). K₁=10^-2, K₂=10^-7, K₃=10^-12.

Indicator:                      pH interval of colour change: 1)thymol-blue                          1.2-2.8 2)methyl-orange                       3.1-4.4 3)methyl-red                            4.2-6.2 4)litmus                                   5.5-7.5 5)phenolphthalein                     8.0-10.1

A buffer will be made that contain Na₂HPO4 and NaH₂PO4. This means that I use K₂, so the c(H⁺) should be around 10^-7. I thought that the right answer is litmus, but it is phenolphthalein. Why?

[Borek]

Will you observe buffer at the end point, or at the middle point of the titration?

[Shadow]

To the end point. Then I have only Na₃PO₄ in the solution. If I use K₃ pH would be 12, but 10.1 is max for phenolphthalein.

[Borek]

What you posted so far suggests you are getting it - at least partially - wrong.

To the end point. Then I have only Na₃PO₄ in the solution.

You can't titrate phosphoric acid to the third end point with 0.1M NaOH. In fact, you can't do it even with a more concentrated base, HPO₄^2- is a too weak acid for that. But before we will get to that there is much more important problem that needs to be sorted out.

If I use K₃ pH would be 12

pH of what will be 12, and why it will be 12? Earlier you wrote

This means that I use K₂, so the c(H⁺) should be around 10^-7

You are consistent, but not correct.

[Shadow]

So it will be titrated to Na₂HPO₄. That's the only substance present in the solution at the equivalence point. pH depends on HPO₄^2-. Kb is bigger than Ka so the hydrolisys will be:
HPO₄^2-+H₂O H₂PO₄^-+OH^-
A buffer is made where K₂ is used. Now I don't know the concentration of HPO₄^2- and H₂PO₄^-. Is the concentration of HPO4^2- two times smaller than the one of NaH₂PO₄ before the titration?

[Borek]

So it will be titrated to Na₂HPO₄. That's the only substance present in the solution at the equivalence point.

This part I can agree with.

pH depends on HPO₄^2-

OK

Kb is bigger than Ka so the hydrolisys will be:
HPO₄^2-+H₂O H₂PO₄^-+OH^-
A buffer is made where K₂ is used. Now I don't know the concentration of HPO₄^2- and H₂PO₄^-. Is the concentration of HPO4^2- two times smaller than the one of NaH₂PO₄ before the titration?

But nothing here makes sense to me.

HPO₄^2- is an amphiprotic substance and pH of its solution can be easily calculated. It is not 12. Judging from what you wrote earlier you think pH at the endpoint equals pKa (you suggested endpoint around pH 7 for pK_a2 and around 12 for pK_a3). That's incorrect - pH=pK_a at middle point of the titration, when concentrations of acid and conjugate base are identical.

[Shadow]

K_a2=c(PO₄^3-)*c(H⁺)/c(HPO₄^2-)
x=square root K_a2*c(HPO₄^2-)
What is the concentraton of HPO₄^2-? 0.05mol/dm³?

[Borek]

x=square root K_a2*c(HPO₄^2-)

No, it is not a correct approach. This is not a way of calculating pH of the amphiprotic substance, this is just a simplified way of finding pH of the weak acid. You even don't know if you can use this approach, as it depends on how far the dissociation goes - and you can't calculate it not knowing the concentration.

To large extent pH of the solution of an amphiprotic salt is not concentration dependent.

[Shadow]

This should be correct: pH=1/2(pK_a2+pK_a3)
pH=9

[Borek]

So if the endpoint is at 9, which indicator is the best?

[Shadow]

9.5 actually, didn't calculate well  . It is phenolphthalein.