[MrHappy0]

I am trying to calculate the pH of a 0.010L 0.681 M HNO3 solution contianing 0.507g of NaClO.

Is this an example of a strong acid where the concentration of H+ is just the concentration of acid and the NaClO is just ignored?

[MrHappy0]

Not sure why no on is replying to my post. Here is my work:

NaClO:

HOCl <--> OCl^-+H⁺

HNO3:

Completely dissociates

---------------------------------------

HOCl Ka= 3.2x10^-8

ka=[H⁺][OCl^-]/[HOCl]=

=([x+H⁺ from HNO3]

• ) / ([F-x]).  

x=3.2x10^-8, pH=7.49

----------------------------------------

I know this is wrong because there is no way the pH is 7.49 because the solution is acidic.

[MrTeo]

Think about this: the HClO that forms from the neutralization of ClO^- won't dissociate if there's some HNO₃ left, as the H₃O⁺ in solution move the reaction equilibrium to the left allowing us to neglect (remember that it's a pretty weak acid) its contribution to the pH. If the HNO₃ is exactly the same as the amount of NaClO put in solution then you simply have to calculate the pH as if there was only a weak acid (HClO). If the HNO₃ is not enough then you'll have to use the buffer equations as you have a mixture of NaClO and HClO. Sorry if I don't check which case is yours but I'm in a hurry... =)

[MrHappy0]

This is confusing. I am not sure which case it is. I believe there are equal concentrations of both but wouldn't I solve it the same way as I did above. If this was not the case and I had to solve it using buffer equations which ones do I use?

[Borek]

This is confusing. I am not sure which case it is.

So check it. It is just a limiting reagent problem.

[MrHappy0]

So since there are equal moles of NaClO and HNO₃ there is 0.681 M of HOCl? Then from there I would do a simple weak acid calculation where Ka=x²/(C-x)?

[Borek]

there is 0.681 M of HOCl? Then from there I would do a simple weak acid calculation where Ka=x²/(C-x)?

That's probably the idea.

[Pradeep]

First check what the reagent in excess is. Nitric or Sodium Oxycholoride? If Nitric larger is in excess Then you may not have to consider  the dissociation of HOCl ( because its very week acid comparing to nitric).

If Sodium Oxycholoride is in excess then you have to use Henderson hasselbalch equation.

[Borek]

First check what the reagent in excess is.

Have you read the thread before posting? She already checked.