[peachgoat950]

I need some help with a question from a major assignment - not the answer, but instructions as to what route to take:

Calculate the molarity, pH, and normality of the resulting solutions when 100.0mL of NaOH with pH of 11.12 is added to 180.0mL of NaOH with pH of 12.89?

Any help is appreciated; thank you!!!!!!!!!!!!!!!!!!!!!!1

[Astrokel]

What's the concentration of OHⁱ(aq) and H+(aq)?

[macman104]

Calculate total moles of OH added to the solution, and calculate the new molarity.

[oxygenfluid]

As macman104 and Astrokel said you need too calculate the total number of moles you have when you get the mix of the 2 solutions. Now, not to give a direct answer but some instrutions, one possible route you can take is:

Calculate the concentration of [OH-] by pOH of both solutions in separate. As you know pH+pOH=14 assuming the temperature its 25 ºC.

After you find the concentration you can calculate the number of moles for each solution.

As you know C=n/V, and you have the number of moles and the total volume, so you can get the molarity of the end solution.

In the end just reverse the 1st step and you get the pH of the new solution.

About the normality this one will be the same of the molarity because you only have one electron permute.

[peachgoat950]

I'm still having trouble figuring out the answer...perhaps because this is the last question of a very long assignment and all the numbers are making my eyes cross 

This is what I have so far...if they aren't right, please just bear with me...and thank you very much for all of your assistance!!

The total volume of the solution is 100.0mL + 180.0mL = 280.0mL

The pOH of the first NaOH is 2.88 (14.00-11.12), the H+ is 7.5 x 10 -12 M
The pOH of the second NaOH is 1.11 (14.00-12.89) H+ is 1.2 x 10 -13 M

I need to find the moles of each in the mixture, but because they are the same composition I'm having trouble....

[oxygenfluid]

ok, you got it almost right, but remind this;

pH= - log[H+] and pOH=-log [OH-], so you can get the [OH-] directly.