Difficulty calculating OH with quadratic formula

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Armature Chemist
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Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

So i have .19g of 100% sodium hydroxide, added to 1L of water, this gives me .00475 moles, no problem there. The PkB of sodium Hydroxide I am given is 13.8.

Now to get the pKa i subtract 14-13.8 which gives me .2

So a Pka .2
pkb 13.8
ok fine

now when I put pka of .200 through my quadratic formula,
Volume
A=1,

ka=B
ka = 0.630957344480193,

initial molarity =C
-molarity X pKa
-.00475 X 0.63095734448019300000 = -2.9970E-03

which when said and done = 4.71477E-03 which when converted to pH is 2.327. Which because it is a base 14-2.327= a pH of 11.673 which is correct.

BUT here is my problem when I reverse it the PKB value of 13.8 in my quad formula, it spits back 0.00000000867653670 which equates to a ph of 5.938
instead of the correct 0.000000000002121 which equates to the ph of 11.673

pKa 13.8
Here is the math
Volume
A=1,

ka=B
ka = 0.00000000000001584893

initial molarity =C
-molarity X pKa
-.00475 X 0.00000000000001584893= -7.5282E-17

which when solved for X = 8.67654E-09 or a pH of 5.938, the X should be at 0.000000000002121 which equates to 11.67

What am I doing wrong?

What I am trying to get is the H+ pka 0.2 to go through my quad formula and read at 4.715E-03 which is a pH of 2.327

and the OH- pkb13.8 to to go through my quad formula to read at 0.000000000002121 which is a pH of 11.673

I can get it to work for the H+ but i can not seem to get the OH- side to work correctly.

Please help me out I am at a complete loss
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ChenBeier
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

I dont understand what you calculating.
19 g NaOH 100% = 19g/40 g/ mol = 0,475 mol
This is in 1 l.
pOH- => - logcOH- = - log0,475 = 0,323
pH = 14-pOH- = 14- 0,323 = 13,67

So what kind of quadratic formula you talking about.
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Hello Chen, it is not 19g as per your calculation, it is .19g as stated in the original post. Which when dissolved in 1 liter of water is .00475 moles.
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

The quadratic formula I am using is pictured in the attachment.

When you use the above values in example 1 where the PKA is. 200 from above, A , B, C, X= the H+ which can easily be switched to pH

However when I use the 13.80 value it is way off as explained above in post 1.
Attachments
Quad formula
Quad formula
Screenshot_20211121-155502_Google.jpg (49.5 KiB) Viewed 133 times
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Chen the problem I find is when dealing with polyprotic acids and bases, to get very accurate numbers the only way to get those really accurate numbers is to run it through a quadriatic formula and then add the results of X

With just using the negative log of the initial pka value (when you have mutiple protic such as EDTA,) it does not provide enough accuracy, hence the use of a quadriatic formula and then adding the results of X for all the protic values, to get your H+ which then you can use to determine the pH value

Now I am trying to do the same using the pkb values of 13.8 to give me the ph of the base without having to subtract the 14. i.e. OH- of 0.000000000002121 which = pH of 11.67345933

There has to be some way to use the pkb values to give you the direct OH- value. something i am missing.

Can I upload an excel sheet here to demonstrate what i am attempting to do?
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

The dot is easy to overlook. :wink: Instead . 19 write .19 or better 0.19 or 0,19. But never mind.
But the calculation goes in is the same way I described.
But I dont understand for what you need a quadratic equation.
What is x,a,b,c in your formula. Its calling midnight formula in mathematics, to solve Parabel equation.

https://en.m.wikipedia.org/wiki/Quadratic_formula

Also what is your context pkb. For instance for OH- its -1,76. Where you get 13.8.

https://www.gr.ch/DE/institutionen/verw ... abelle.pdf
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Hey Chen,

The 15.7 for NAOH, is incorrect in some chemistry books due to the water constant, once the water constant is taken into consideration the pkb is at 14.8 in most chemistry books.

https://onlinelibrary.wiley.com/doi/abs ... .201300321

https://actingcolleges.org/library/acti ... ue-of-naoh

See the Bordwell organic PKA tables say water has a pH of 15.7 https://organicchemistrydata.org/hansre ... urces/pka/

When in most tables it has a pH of 14, pdf download
https://www.nku.edu/~russellk/courses/c ... -short.pdf

Hence sodium hydroxide would have pkb of 13.8 vs a higher value, as I understand it due to the water discrepancies.

https://pubs.acs.org/doi/pdf/10.1021/ac ... ed.6b00623#

As well the pka value I was given to use was 13.8.
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Hey Chen

I am using the quad for 2 reasons.
1. Accuracy
2. In the US it is commonly taught,
3. Gives specific values of speciation

This video https://youtu.be/nXugjkSQtTI at time 9:06 shows the quadratic used for polyprotic acid namely H3PO4.

Like I said the only problem I am having is on the flipside, I can calculate the H+ no problem for any acids, But I can not, using g the Pkb value get it to spit out the right OH value.
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

Now I understood what you are talking about. But you mixed strong acid/ base with the weak polyprotic once like HSO4-, H3PO4 and its dissociated. Of course here we have to consider all H+ especially dilution is high and also dissociation is high.
But for one protic acid like HCl or base NaOH these calculation are not necessary.

In the midnight formula

What is in your example a,b and c. Can you clarify. I get dizzy by numbers like 0.000000032 etc. Better is use Exponent 3.2 *10^-8
So what is ka, kb in your formula.
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Correct for monoprotics its not needed but for multiprotic it very much is needed, also I have found much more accuracy in the quad formula, for monoprotics, than using the simple version, on some mono protics I have seen as much as a .03 to .15 variation using simple methods vs the quad. Personally I like the quad for accuracy, however it has its problems too like calculating the OH in an acid...

So the question still stands, how do I get the quad formula to calculate the OH- from the 13.8

So when I put the H+ of .2 in, the formula, I get the appropriate H+ to calculate the pH which is like 2.37?? Subtract that from 14 we get the 11.6ph

But I am trying to use the 13.8 to get the OH- which can be directly converted by -log of ... without the subtraction.

So how do I do this.

With the quad formula I can get the H+ to come out correct for acids, but not the OH-,

With the quad formula for bases I can get the OH- to calculate correct, but not the H+...

What am I doing wrong that I can get one but not the other to come out right?
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

I think the problem is the 0,0475 moles. This has also converted to H+ isn't it.
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

I will post my excel sheet on one of the excel help forums later today, so you can see more clearly what I am trying to do.

There has to be a way to use the pkb to calculate the OH- using a quadratic formula. I just don't see it, it has to be a simple error in my formula or a simple math error.

I am trying to get the quad to do the work instead of adding additional steps like subtracting....
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

pKb = 13,8 pKa = 0,2
c OH- = 0,00475 mol/l , c H+ = 2,11 * 10^-12 mol/l
Calculate with this one
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Re: Difficulty calculating OH with quadratic formula

Post by Armature Chemist »

Chen Here is the Screen shot of the excel sheet to better illustrate what I am meaning by I cant seem to calculate the OH- correctly.

Now keep in mind that this is ONLY for a mono protic base, as we have discussed earlier, I am making this to calculate Polyprotic acids and bases, which in such case MUST go through the Quad formula, whether it is a mono or polyprotic acid, the quadriatic formula (in theory) should give you the correct values, I got them as you can see working for the H+ but can not figure out for the life of me why they are not coming up correctly for the OH-

I will upload this excel sheet in a few moments and post webpage and PM you the site
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Re: Difficulty calculating OH with quadratic formula

Post by ChenBeier »

Did you read what I wrote above. If you use kb then c OH- has to be used. If you use ka then c H+. I think that is the key. First figure out what is the right calculation. Excel is only a tool, what will not solve the problem.

initial molarity =C
-molarity X pKa
-.00475 X 0.63095734448019300000 = -2.9970E-03

initial molarity =C
-molarity X pKa
-.00475 X 0.00000000000001584893= -7.5282E-17

In both cases c OH- = 0,0475 mol/l is used
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