[Edher]

Saludos,

I'm having trouble starting this problem, the book doesn't offer similar examples.

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

I'm utilizing the ICE method by setting up the reaction in the following way,

HNO3 + KOH <--> H20 + KNO3

However, I don't know what to do with the amount of the solutions (mL.) Could I get a hint?

Thank you,
Edher

[jdurg]

Well, here's a couple of hints.  HNO₃ is a strong acid, and KOH is a strong base.  So when in solution, they fully ionize.  Therefore, you just need to figure out which one you have more of.  They give you the molarity of each which means number of moles per liter, and they give you a volume in terms of mL.  So once you figure out the number of moles of HNO₃ and KOH that you have, you'll be able to figure out which one is in excess.  The excess will be all that's left, in terms of H+ and OH-, in the final solution which is made by completely mixing the two volumes.  

[Edher]

This is what I did now,

    I found out the number of moles of each substance by creating two ratios (one for each substance.)
    Then I subtracted the amount of moles of each substance to determine which is the excess. (Since the excess would be the amount of moles that do not ionize to make either KNO3 or H20.)
    In this case, there was a 8.28x10^-5 moles excess of HNO3.
  It just hit me right now,
      To find the molarity, I have to take the number of moles in excess and divide by the net volume. Then take the negative log and viola. (pH=2.7)

However, why am I using the total volume of both solutions, I mean, not all (0.0402 L) of solution is water, there's also K+ and NO3- ions dissolved in solution. They don't combine with the H20 molecules but they still take up space right? So they would take up part of the (0.0402 L) of solution. Wouldn't I then have to subtract the volume they take up from the total solution and then use that amount of water?

Edher

[jdurg]

Your answer is correct.  The pH of the solution would be 2.68.  The amount of water generated and the space the ions take up is completely negligible.  Their volume plays no part in determining the overall pH of the solution, especially to the accuracy of the measurements you have.  For example, in this case you have .0060676 moles of water being formed.  At ~18g/mole, that's ~0.109 grams of water.  With a density of about one gram per mL, that gives you an added volume of about 0.1 mL.  That is completely negligible in your overall calculations.