Difficulty calculating OH with quadratic formula

[Armature Chemist]

Chen, I hear what you are saying, I get it, however I am NOT seeing it.
Are you saying the problem is in C value in (S₇) with the initial molarity? or are you saying that the molarity in C₁₆ is wrong?

This is the formula that I am using for the base OH

(M₁₂) Pkb₁ 13.8
(P₁₂) kb 1 1.5849E-14
(c16) total moles .00475 ( g of substance .19 / molar mass 40)
(C₁₃) volume 1L

so when put into the quad,
(S₅) a= volume (1)
(S₆) b = pkb (1.5849E-14)
(S₇) c= -7.5282 (-.00475* 1.5849E-14)

which when calculated is returning 8.7E-09 which relates to pH of 8.06.

You are saying if I understand you correctly there is a problem with the molarity
can you please point out which cell you are referencing, so that I can fix the formula to that cell cause I am just no seeing what you are referring to in the formula below the red box.

[ChenBeier]

Let say in this way you have a NaOH solution containing 0,19 g = 0,00475 mol NaOH = OH- , in this solution we have also H+ .

So if you have kb then use OH-
If you use ka then instead H+ has to be used.

It's similar pH = - logcH+, but pOH = -logcOH-

You can not use the same number 0,00475 for cH+ and cOH- in same time.

The cH+ in a NaOH solution is very low. cH = 10^(-pH) = 10^(-11,67)= 2,11*10^(-12) mol/l

[Armature Chemist]

I am using (am I not?) Pka and Ka for H+ and Pkb Kb for OH- please see new attachment.

Are you suggesting that it is the molar of OH- i.e. .00475 in C₁₇ and then subtract the molar of the H+ in V₃ ??

[Armature Chemist]

Chen I get what you are saying, if the OH is say 10 and the H is 4, 10+4= 14
if you know that the sum is 14 and know that the OH is 10 14-10= 4 I get that

what I dont get is where (what cell) am I making the mistake, and HOW do I fix it.

am I simply taking .00475 (OH) and subtracting the .00471 = .00004 moles and run that through the quad??? this is what I dont get, WHERE am I making the mistake, and What is the solution?

You can see why I used Armature right, the information is circling around my head, but until some one points me at a target, it keeps just spinning lol.

[ChenBeier]

In the box R 14 the - moles has to be used as H+. You used OH-. Probably also the equilibrium of water has to be considered but I didn't checked it.

Also in the box X 3 above the result ist not pH its pOH. So there is some mixed up generally.

[Armature Chemist]

Hey Chen
Lets go back to basics for a moment.
When we look up the PK value for hydroxide it says it is 13.8
as it is a base, the 13.8 would be the PKB value correct?
The PKB value there fore measures pOH (OH-) correct?
from the PKB values we can then solve for the KB values.

If we take 14 and subtract 13.8, we get .2
Now as .2 is the difference between the H+ (14) and the OH- 13.8 the resulting .20 is measuring the PKA value correct?
Thus the PKA value would be measuring the H+ correct?

In which case in the first formula example above the red line I am using the PKA and Ka values, which would give me the H+

In the second case I am using the PKB and the KB values which should give me the OH- values

I am not using the same numbers for both sets, at least not that I am recognizing.

[ChenBeier]

That is all OK, but you use in both cases the same concentration for OH- and for H+ what was 0,00475 mol. In one case you get the right numbers in the other case not. The H+ concentration is different.

cOH- of 0,00475 refers to pOH 2,323 or pH 11,676
This alkaline solution contain H+ ions as well.
pH of 11,676 refers to 2,1*10^-12 mol/l H+

https://www.thoughtco.com/ph-pka-ka-pkb ... ed-4027791

[Armature Chemist]

Hey Chen here is the link to the actual excel sheet, https://www.excelforum.com/excel-genera ... error.html

I am looking at your last comment and am trying to process it.

[Armature Chemist]

Hey Chen, You mentioned molarity, I have done some experimenting by manually entering numbers in C₁₄. The only way I can get it to work is by putting the molarity of 0.00000000028596459

However I fail to see where this molarity value is coming from please see the attached photo.

Where is this value coming from as it is not from the .19/ .000475 Please advise.

[ChenBeier]

I read the thread in the excel forum.
I think there is already some confusion.
They talking about add NaOH and pH get acidic. That is wrong. The mistake is in your sheet you claiming for pH, but it isn't . The 2.32 is the pOH.
Also there is a mixup between pka, pkb pH and pOH

There are two equations. pKa + pkb = 14
and cH+ * cOH- = 10^-14 in logarithm pH + pOH = 14

What is pKa and pKb

HA + H₂O => H₃O+ + A-

ka =cH₃O+*cA-/cHA => pKa = pH +log(cA-/cHA)

MeOH + H₂O => Me(H₂O)+ + OH-

kb = Me+ *OH-/MeOH => pkb = pOH- + log(Me+/MeOH)


You have cOH- = 0,00475 mol/l how much is cH+

With cH+ * cOH- = 10^-14 => cH+ = 2,1 *10^-12

pkb =13,8 so pks = 0,2.

[Armature Chemist]

Hello Chen

I have changed the PKA and PKB values and have created 2 different formulas to show the H+ on one and the OH- on the other for the same compound to avoid any confusion on the values.

I still cant get this to match the Chembuddy ph, nor the Curtipot pH calculator on the second formula.

I am using the pka/ka values to calculate the H+, I am using the pkb/kb values for the OH- but i still cant get it.

[ChenBeier]

The same error. Mix of H+ and OH-.

[rajsarvan708]

When you say difficulty calculating OH with quadratic formula, are you working on a chemistry problem (finding hydroxide ion concentration
[
𝑂
𝐻
−
]
[OH
−
] using the quadratic formula from an equilibrium expression), or a math/geometry problem (maybe finding a line segment
𝑂
𝐻
OH in a triangle using quadratic equations)?

If it’s chemistry, here’s the usual situation:

You have a weak base (or weak acid), and you set up the equilibrium expression.

The concentration of
𝑂
𝐻
−
OH
−
often leads to a quadratic equation of the form:

𝑥
2
+
𝑏
𝑥
+
𝑐
=
0
x
2
+bx+c=0

where
𝑥
=
[
𝑂
𝐻
−
]
x=[OH
−
].

Then you apply the quadratic formula:

𝑥
=
−
𝑏
±
𝑏
2
−
4
𝑎
𝑐
2
𝑎
x=
2a
−b±
b
2
−4ac
​

​


and choose the physically meaningful root (since concentrations can’t be negative).

If it’s geometry, then “OH” might mean a line segment (like altitude or median) in a triangle, and you’re trying to calculate it with a quadratic relation.

[ChenBeier]

Please Format your Text, nobody can read it.
And don't answer 4 year old post.