Chemistry Forum

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 a...

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 ...

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.

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 subtra...

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 = .0...

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 ??

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) tot...

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 ...

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...

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...

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 flipsi...

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/10.1002/hlca.201300321 https://actingcolleges.org/library/acting-questi...

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 mut...

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.

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.