[slayer]

If you mix 5.00 mol of NaBrO with 2.50 mol of HBr in 10.0L of water, what will be the pH of the resulting solution?


I was waaaaay off when i did this problem. This is how I tried to solve it.

since the solution = 10L, I used the mols of each substance and divided it by 5.0L, assuming that the 10L consisted of equal portions of NaBrO and HBr

therefore, 2.50mol HBr / 5.0L = 0.5M
and 5.00mol NaBrO / 5.0L = 1M
I think i was way off here too. but i couldnt find another way to determe the starting molarity of the
reactants.

this is the equilibrium reaction that I decided to construct:
BrO^- (aq) + H₂O (l) <----> HBrO (aq) + OH^-

I then created an ice table
[1; -; <--> 0.5; 0]
[-x; -; <--> -x; +x]
[1-x; -; <--> 0.5 -x; x]

0.5 - x = x
0.5 = 2x
x = 0.25
-log [0.25] = .602 = pOH

pH = 14 - pOH
therefore, pH = 14 - .602
pH = 13.4

this is so wrong. but i dont know where I made my mistakes.
can someone please help me?

[Borek]

You reaction equation describes just solution of salt of weak acid. You have strong acid in the solution as well.

This will be a buffer solution.

[AWK]

Assuming HBrO is stable, and dilution of buffer does not matter, you have 2.5 moles of NaBrO and HBrO in this case pH is approx. equal pKa of this acid

[slayer]

so with the equation

pH = pKa + log [[A-]/[HA]]

I am supposed to figure it out. But I dont know what Im supposed to do with this equation. I dont know what my molarities are. I cant just take the given 2.0moles of HBr and divide it by 10 liters of water, right?

and i assume with the pKa, Im supposed to find that on the pKa/ Ka appendix.

How do I construct my equilibrium reaction?

[Borek]

I dont know what my molarities are.

Assume that protonation of BrO^- was complete. That's the way you do it. See pH of uffers for more detailed discussion.

[slayer]

So borek, what youre saying is that there is 100% dissociation in this buffer?

if there is 100% dissociation, and my concentrations are 1:1, pH = pKa?

[Borek]

I would not call it 100% dissociation, although I have no better name for what's happening here. More like no dissociation at all - just what was neutralized stays neutralized.