[Edher]

Saludos,

I'm having trouble solving this problem:

What is the pH of the solution obtained by mixing 24.80 mL of 0.248 M HNO₃ and 15.40 mL of 0.394 KOH?

I know that in order to obtain the pH I must determined the conentration of H₃O in solution. However, I believe this is the equation of the reaction:

HNO₃ + KOH <> H₂O + KNO₃

How can I determin the concentration of H₃O when it is not one of the products?
Is my equation incorrect?

Thank your in advance for your attention and your help,
Edher

[Edher]

I think I have figured out a new approach, instead of having one equation, perhaps I could use two. In other words one for HNO₃ and another for KOH.

HNO₃ + H₂O <> H₃O + OH-

and

KOH + H₂O <> H₃O + OH-

and then just find the H₃O concentration on each one and then add them. Though I tried that and I didn't get the same answer in the book. Quite close though.

Am I on to something here?

Edher

[madscientist]

Hi Edher,

I think this is how its calculated:

pH = pka + log ([conjugate base] / [acid])

HNO3 is acidic, KOH = conjugate base

the pka value for HNO3 (nitric acid) is:  

pka=-log ka=-1.30

and the molar concentrations are:

[Potasium Hydroxide] = 0.394 M

[Nitric acid] = 0.248 M

so:

pH = (-1.30) + log(0.394M / 0.248M)

     = (-1.30) + log(1.589)

     = -1.098

     = -1.1

extremely acidic!


cheers,

madscientist :albert:

[Borek]

HNO₃ + KOH <> H₂O + KNO₃

Correct. One of the substrates is in excess - calculate which one. What is left will define pH (or pOH) of the solution.

That's assuming that both KOH and HNO₃ are strong (satisfied precisely enough).

[Borek]

pH = pka + log ([conjugate base] / [acid])

It is called Henderson-Hasslebalch equation.

HNO3 is acidic, KOH = conjugate base

No. HNO₃ is acid, but its conjugated base is NO₃^-. Read about Bronsted-Lowry theory.

the pka value for HNO3 (nitric acid) is:  

pka=-log ka=-1.30

Closer to -1 as far as I know.

As you have built rest of your post on the misunderstanding of B-L theory, it is completely wrong.

Solution has pH below 3, but is not as acidic as you think.

[arnyk]

Ok stop me if I'm wrong since I'm pretty rusty but here's a more simpler way:

24.80 mL of 0.248 M HNO3 and 15.40 mL of 0.394 KOH?

HNO3 + KOH --> KNO3 + H2O

1:1:1:1

n (HNO3) = 6.15e-3 mol
n (KOH) = 6.07e-3 mol

excess HNO3 = n(HNO3) - n(KOH)
                    = 8.28e-5 mol

c = n/V
   = 8.28e-5/0.0402
   = 2.06e-3 M

pH = -log[2.06e-3]
pH = 2.69

 

[Edher]

Well, according to the book the answer is pH=2.7.

Some of your methods seem foreign to me, specially this Henderson-Hasslebalch equation. I guess they are more advance since right now I'm learning the Bronsted-Lowry theory. The reply that I can comprehend the most (since I attempted something similar) was ARNYK'S. How did you calculate the excess though? I know how you got the number of moles, from there on I got lost.

kudos on your picture madscientist.

Edher

[arnyk]

Ok fixed it up it here, forgot to divide by the total volume.

There are more advanced ways to do it, but I figured that'd be good enough.

[Edher]

Excellent, I got it now. Thanks a lot everyone, you've been of great help.

Edher

[madscientist]

Sorry, just tryin to help but maybe a little out of my league

cheers,

madscientist
(:stupid:aka NUMBNUT )

[Edher]

Your participation, your effort and your presence  (specially with that cartoon) is highly appriciated madscientist.

[Borek]

The reply that I can comprehend the most (since I attempted something similar) was ARNYK'S.

Funny - he did exactly what I told you to do - he calculated which reagent is in excess

[rcoleman]

Funny - he did exactly what I told you to do - he calculated which reagent is in excess

But aren't you supposed to be mixing the two solutions (I'm currently working on a similar problem)? Why would you need to calculate the limiting reactant when you're mixing both of the solutions together anyway? I get what arnyk did, I just don't get the thought process behind it or what you are saying. Sorry if I'm coming off a little slow, I just want to understand for future reference, i.e. a test.

Hmm, I think I understand. When you mix an acid and a base they are going to neutralize, so whichever molecule has more moles\substance present than the other one is going to be left over, so this is where you calculate the pH. Am I correct?

[Borek]

Hmm, I think I understand. When you mix an acid and a base they are going to neutralize, so whichever molecule has more moles\substance present than the other one is going to be left over, so this is where you calculate the pH. Am I correct?

Exactly  

That's not as easy in the case of weak acids and bases, but in the case of strong ones - they react till one of them is used away. What is left changes pH of the solution.