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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: uncreative on December 21, 2016, 11:28:48 AM

Title: pH exercise
Post by: uncreative on December 21, 2016, 11:28:48 AM
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?

Title: Re: pH exercise
Post by: Borek on December 21, 2016, 12:01:55 PM
Would knowing the base concentration help?
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 12:07:21 PM
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?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 01:22:54 PM
From pH of weak base with known Ka or Kb (pKa or pKb) 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
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 01:38:54 PM
Ok that's true i can simplify the equation,, but still how do I know that [OH-]=[B+]?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 01:59:22 PM
Quote
but still how do I know that [OH-]=[B+]?
From reaction of dissociation (neglecting autodissociation of water)
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 02:31:01 PM
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?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 02:35:34 PM
Two solution are mixed (equal volumes)
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 02:40:24 PM
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-]?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 02:48:37 PM
Quote
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.
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 02:57:55 PM
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
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 03:06:41 PM
Quote
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.
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 03:09:59 PM
Quote from: AWK on December 21, 2016, 03:06:41 PM
[quote And concentration of hydroxide ions depends on concentration of ammonia.

what is the dependance between these 2?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 03:11:04 PM
http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
Title: Re: pH exercise
Post by: uncreative on December 21, 2016, 03:29:22 PM
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?
Title: Re: pH exercise
Post by: AWK on December 21, 2016, 03:51:33 PM
It is not true for weak bases, as in this case.
Title: Re: pH exercise
Post by: AWK on December 22, 2016, 06:55:03 AM
Read Chembuddy lecture carefully (link is given twice). Check applicability of equations 8.8 and 8.12 to your problem (use Kb instead of Ka and OH- instead of H+).
Calculate ci for both solution (check also table Results of pH calculation of acetic acid). Then for then mean value of c (=> equal volumes) calculate OH-, pOH and finally pH.
The final pH should be between values given for you solutions, but very close to one of these values.
Title: Re: pH exercise
Post by: docnet on December 23, 2016, 02:14:50 AM
Use the Henderson Hasselbalch equation: pOH = -log(Kb) + log (B+/BOH) and use ratios to find the pH of the final solution, taking into account that equal volumes of the two solutions are mixed!

You might not this here, but the following equations are useful for pH problems.
[H+] = 10-pH
[OH-] = 10-pOH

I hope this helped at least a little bit. Have a nice day!
Title: Re: pH exercise
Post by: AWK on December 23, 2016, 04:29:58 AM
Quote from: docnet on December 23, 2016, 02:14:50 AM
Use the Henderson Hasselbalch equation
H-H equation is useless for single weak acid or base.