[uncreative]

Calculate the pH that you obtain mixing same volumes of 2 solutions of the same base(Kb = 1.85 * 10^5), one having pH=8.65 and the other one having pH=9.25.

Well, I don't understand the problem. I mean, I know the formula for pH=-log[H+]

and that of Kb, but how can I find the final pH?

[Borek]

Would knowing the base concentration help?

[uncreative]

Kb= ([B+][OH-])/[BOH]   right?


I can find [OH-] using the pH but then I don't know 2 things in that equation. Where am I wrong?

[AWK]

From pH of weak base with known K_a or K_b (pK_a or pK_b) you can calculate easily concentration this base in both solutions. after mixing you can find a new pH.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[uncreative]

Ok that's true i can simplify the equation,, but still how do I know that [OH-]=[B+]?

[AWK]

but still how do I know that [OH-]=[B+]?

From reaction of dissociation (neglecting autodissociation of water)

[uncreative]

so, after i  find [BOH] for both solutions, what is my next step? I mean, I know:

1 [OH-] for solution 1
2 [OH-] for solution 2
3 [BOH] for solution 1
4 [BOH] for solution 2


I can't properly figure out how I find the new pH.

I need the final [OH-], right?

[AWK]

Two solution are mixed (equal volumes)

[uncreative]

Yes, but the volume is unknown, I need the molar concentration of BOH and if I knew the number of moles i could find the volume, but being it unknown, how do I find final [OH-]?

[AWK]

Calculate the pH that you obtain mixing same volumes of 2 solutions

Volumes are known! You may take any two the same volumes (eg 10+10, 100+100, 500+500). Moreover, there is shortcut - the same result is obtained from arithmetic mean of both concentration (neglecting volumes) for this data.

[uncreative]

I'm so sorry, but:

final [OH] = total moles (OH)  /  2V;


solution 1: moles (OH)= [OH]*V;
solution 2: moles (OH)= [OH]*V;


can you tell me why I need to use [BOH]? This is what I don't understand.


P.S.

By "I'm sorry", I mean sorry I keep disturbing you

[AWK]

final [OH] = total moles (OH)  /  2V;

This is not true. As number [OH^-]is not equal to [NH₃]
You mix solutions of NH₃ obtaining a new concentration of ammonia. And concentration of hydroxide ions depends on concentration of ammonia.

[uncreative]

[quote And concentration of hydroxide ions depends on concentration of ammonia.

what is the dependance between these 2?

[AWK]

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[uncreative]

But isn't true that

final n moles (OH) = n moles (OH from sol 1) + n moles (OH from sol 2)?

or is this also not true?