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?