ph Calculations....of NH4H2PO4 and KH2PO4
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#1
May 31st, 2005, 15:15
 angelguy Junior Member Join Date: May 2005 Posts: 3
ph Calculations....of NH4H2PO4 and KH2PO4

Hi , This is Ashmit

I have been solving this problem since almost a week now, however, this stuff is getting more and more complicated.

The object is to determine the ph of KH2PO4 and NH4H2PO4 (mono basic) . I have done lab on both of these by taking known amount of salts and got 4.5 for KH2PO4 and 4.4 for other one.In this lab, I have taken 0.1156M of these compounds and dissolved it in 1 liter of water(pH 7) and then measured the pH. @ 25°C

These days my major concentration has been in the calculation of KH2PO4 since i think it will be simpler to deal with. I have assumed the complete solubility of KH2PO4 . and the key ion affecting the pH is dihygrogen phosphate H2PO4(-) . This gets dissociated into HPO4(--) and proton. with pKa value of 7.2. Although it further dissociate but not a big deal since pKa value of 12.35. I might have tried maximum possible methods I know but I cant get the close agreement with the lab value.
I have also used Henderson equation and it yield me the answer of 4.2

This is what i did,
pH = Pka + log([HPO(--)]/[H2PO4(-)] and Pka is 7.2 also I hv assumed that the conc of H2PO4(-) is 0.1156 M by stoichiometry.
and thereby conputed the value of HPO(--) by this formula . .
ka = x^2/(0.1156 - x) and got x = 9.89 * 10^(-5) aproxx which is the value of HPO(--).

Kindly guide me in this situation and also give me suggestion on how should i approach NH4H2PO4. situation. I will really appreciate if any one of you guys help me out since i am gonna get crazy doing one thingy over and over again. :?: Also had a confusion wrt considering NH4(+) in other situation. Please tell me mistakes in calculation.. which i m sure i did.

thanks a bunch..

Ashmit
#2
May 31st, 2005, 18:02
 Borek Member Join Date: May 2005 Posts: 32

Check

http://www.odu.edu/sci/xu/chapter11slide.pdf

for calculation of KH2PO4 pH.

Amonium dihydrogen phosphate is pretty similar. NH4+ is too weak acid to change anything substantially.

In such concentrated solution classic calculation will not yield proper results - activity coefficients are too large!

Check my BATE for fast pH calculation.

Best,
Borek
#3
June 1st, 2005, 12:34
 angelguy Junior Member Join Date: May 2005 Posts: 3

I have already downloaded your BATE and tried working the same compounds( KH2PO4) in it, however, I got pH = 4.07 for 0.11565M concentration. This value is pretty off from the experimental results pf 4.5 . I was kinda wondering , like did u guys considered the activity coefficient of all the ions in the software ?? Because after considering the activity coefficient of the pH my answer comes like 4.35 which is much closer to experimental result 4.5

I have computed the activity coef of these ions by using debye-huckel equation. Kindly guide me if the approach is correct ? then computed the other ka which is used for the actual calculation.

I really cant figure out why the pH(exp) of KH2PO4 and NH4H2PO4 are different for same concentration of correspoding salts. ?? because i was really confused, i performed the lab again doing the same thing but i got the same results. Mathemethically, I got the same answer for the NH4H2PO4 pH as KH2PO4, if I were to neglect the effect of NH4+. which is quite obvious. ..

the reference you provided was extremely helpful thanks a lot for that. It helps me a lot in carrying calculations. But I cant really find the actually calculation of KH2PO4. I got the one with phophoric acid..which takes H3PO4 + H2O ---------------- H2PO(-) + H30(+) in account .. which i think i dunt need it. because my salts dissociated directly into H2PO(-) ion . Also the Ka value of this predominate the actually caluclations..

Finally, Also my ultimate aim is to mix all different salts and then have to determine the ionic species at different pHs in a container mathemetically. I can name some of the important salts in the container.. 1) NH4H2PO4, MgSO4, NaCl, CaCl2 and so on.. I will be thankfull to you if you can also guide me about the approach to be taken to complete this query. Hoping for your favorable response again..

Thanks a lots

Ashmit
#4
June 1st, 2005, 17:54
 Borek Member Join Date: May 2005 Posts: 32

Quote:
 Originally Posted by angelguy I have already downloaded your BATE and tried working the same compounds( KH2PO4) in it, however, I got pH = 4.07 for 0.11565M concentration. This value is pretty off from the experimental results pf 4.5 . I was kinda wondering , like did u guys considered the activity coefficient of all the ions in the software ?? Because after considering the activity coefficient of the pH my answer comes like 4.35 which is much closer to experimental result 4.5
You must have done something wrong. Start BATE and select phosphoric acid and potassium hydroxide as present in the solution. Enter 0.1156 for both concentrations - this is equivalent of KH2PO4. pH is calculated by the program as 4.69. There is a drop down list at the bottom of the dialog window - by default it says "ignore I". Select "calculate I" - and the pH calculated changes to 4.44 (note ionic strength of the solution right to the drop down list).

Quote:
 I really cant figure out why the pH(exp) of KH2PO4 and NH4H2PO4 are different for same concentration of correspoding salts. ?? because i was really confused, i performed the lab again doing the same thing but i got the same results. Mathemethically, I got the same answer for the NH4H2PO4 pH as KH2PO4, if I were to neglect the effect of NH4+. which is quite obvious. ..
It is enough if the salts used for preparation of solutions are not 100% pure - the pH measurements can be repeatable, but slightly different from expected. 0.1 pH unit is not a large difference (well, sometimes it is, I don't know what pehameter you used - note that if you have measured pH with one decimal digit accuracy it is enough if both solutions pH differ by 0.01 to display different values).

Quote:
 the reference you provided was extremely helpful thanks a lot for that. It helps me a lot in carrying calculations. But I cant really find the actually calculation of KH2PO4.
Look for pH = (pKa1 + pKa2)/2, perhaps it was presented for diprotic acid, not for triprotic - but it is the same approach. Third proton in phosphoric acid is nailed :wink:

Quote:
 Finally, Also my ultimate aim is to mix all different salts and then have to determine the ionic species at different pHs in a container mathemetically. I can name some of the important salts in the container.. 1) NH4H2PO4, MgSO4, NaCl, CaCl2 and so on.. I will be thankfull to you if you can also guide me about the approach to be taken to complete this query. Hoping for your favorable response again..
That's not so easy - there is no universal method that can be always safely used. In general, you should write down all equations describing the solution (mass balances, charge balance, water ioniozation and all dissociatian equilibria) ang go from there - some sums can be approximated by neglecting smaller values. Some substances can be treated as dissociated 100%. But it is always intuition and every assumption have to be later checked whether it holds.

Best,
Borek
#5
June 2nd, 2005, 14:15
 angelguy Junior Member Join Date: May 2005 Posts: 3

hey .. I dunt know how to express but i am extremely thankful to you...

Actually, I am using this in determining the ionic concentration of ionic species in fermentation media. therefore, I need to compute the different concentration at the set pH. Well, this is just the start ... and which why i started up computing the pH of the simple systems.

Secondly, I did quite figure how mathemetically you ended up getting the value of 4.44 and also was not quite sure how did u get 4.69 itself.. Could just gimme the hint like which equation u used for computing the pH without the activity coefficient(I) and with it. I will be extremely glad if you could help me in this situation.

Also didnt understand the derivation of the diprotic equation, like pH = Pk1 + Pk2. I have a firm believe that the phosphoric acid will not interfere my case . Thanks again to make me understand that acid do interfere in situation..I hope you will help me solve these two questions ..

You calculator is extremely help it made my life a a lot easy.. Thanks to trial version.

Thanks a lot

Ashmit
#6
June 2nd, 2005, 17:30
 Borek Member Join Date: May 2005 Posts: 32

Quote:
 Originally Posted by angelguy Could just gimme the hint like which equation u used for computing the pH without the activity coefficient(I) and with it. I will be extremely glad if you could help me in this situation.
http://www.chembuddy.com/?left=BATE&right=phcalculation

I doubt it will be helpfull. BATE does a brute-force numerical attack at the system, you will be not able to repeat it by hand. Hey, that's what computers are for

Ionic strength and activity coefficient equations are also there.

If you have more questions check my FAQ section for chemicalforums link - I am more frequent there as I am one of moderators.

Best,
Borek

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