[angelguy]

hi Borek,
This is Ashmit

Actually i have some questions about BATE pH calculator.

As you might be knowing the pH computed by BATE was 4.69 without the activity coefficient and 4.44 with activity coefficient..
Well, when I trying to understand the actual calculation behind it. I figured out in equilibrium composition section that the pH computed was 4.31. Could you explain what is the difference between these pHs. i.e 4.69 and 4.31.

When the H conc(computed by 4.31)  multiplied by the activity coefficient of h3O+ ions gives pH of 4.44.  Could you please explain me by which equations other Ph like (4.31) comes.. and why did we used phosporic and pottasium hydroxide instead of the actual ions.(H2PO4- and H3o+)??

Actually, I tried computing the complete thingy by activity coefficient. Let me give the detail of it,,
I computed the Ka'(Ka with activity coefficient ) by multipying Ka (original, 7.2) with the activity coefficents of H2Po4(-), HPO(2-), H3O+. then I  tried to pluggin the equilibium conc. in H-H equation. But was confused wat Ka in henderson equation to be used (original or the Ka').
My situation is,  if i use the original Ka i can get 4.31( n my prob is sloved )  and if i do it with the original ka , I end up gettin 3.8 .  

Please help me by giving some more detail about the BATE calculations and also the senario i mentined above.

Thanking you in advance.

Ashmit

[Borek]

To all: this is continuation of discussion started on another forum server, I asked angelguy to move here

As you might be knowing the pH computed by BATE was 4.69 without the activity coefficient and 4.44 with activity coefficient..
Well, when I trying to understand the actual calculation behind it. I figured out in equilibrium composition section that the pH computed was 4.31. Could you explain what is the difference between these pHs. i.e 4.69 and 4.31.

4.31 is -logarithm of just concentration. Remember, that pH is defined as activity scale. 4.31 is a result that you will get measuring H⁺ concentration with spectroscopic methods (if they exist, can't remember any suitable).

When the H conc(computed by 4.31)  multiplied by the activity coefficient of h3O+ ions gives pH of 4.44.  Could you please explain me by which equations other Ph like (4.31) comes..

4.31 is calculated exactly opposite way - first, acitivity is calculated (numerically). Then, activity is divided by the activity coefficients to give real concentration.

and why did we used phosporic and pottasium hydroxide instead of the actual ions.(H2PO4- and H3o+)??

To overcome limitations of the user interface - at present it doesn't allow to enter salt concentration, but it allows to create any mixture of acid and base. Potassium dihydrogen phosphate is nothing else then equimolar mixture of phosphoric acid and potassium hydroxide (once the neutralization reaction ends, but it is very fast).

Actually, I tried computing the complete thingy by activity coefficient. Let me give the detail of it,,
I computed the Ka'(Ka with activity coefficient ) by multipying Ka (original, 7.2) with the activity coefficents of H2Po4(-), HPO(2-), H3O+.

Don't ever do it. Ka is defined and measured using activities, so there is no need to modify its value. Instead, do the calculations using activities.

It is not easy to calculate activity coefficients for acid/base equilibrium by hand. Why? Because that's iterative process. You start with classical calculation of concentrations, then you calculate ionic strength and acitivity coefficients, you put them into Ka equation, you calculate new concentrations and you repeat the process untill the results of two consecutive steps differ less then predefined treshold.

In your case (KH₂PO₄) situation is rather simple, as you may start calculations assuming known concentrations of K⁺ and H₂PO₄^- and calculating ionic strength and activity coefficients using these values. H₂PO₄^- dissociation is so weak that it will not change ionic strength of the solution much (abundance of K⁺ and H₂PO₄^- creates so called ionic strength buffer). Thus one step of calculations should be sufficient.

[angelguy]

Hey , This is Ashmit,

I further have few question regarding our previous discussion.

1). I really can not figure out wat u mean in last reply. Like, if 4.31 were to be the -log of H, then wat do we mean by the pH of 4.69?

2). Also I calculated the ph = 4.69 by averaging the two Pkas, also suceeded in computing the values of activity coefficient of by using Dybye-Hukel Equation in calculation center . This value hold with value computed by BATE. However, if i use activity co-ef, I cant get the value of 4.45 by using 4.69 as pH. Instead, the pH increases value increase above 4.69 . Well, I can get the pH of 4.45 by multipying activity coefficient by the hydrogen conc. obtained by 4.31. I guess for that I should understand the numerical method of BATE obtained the values of 4.31. Therefore, In last reply u mentioned some numerical method for obtaining. Could you please help me find that method. or provided me some reference for the same

3). Also in the formula for diprotic acid, ( pH = 1/2(pK1 + pK2)) . I cant figure out the concentration of H2A and A- are assumed to be same

Thanking you in an anticipation.

Ashish Chitalia

[Borek]

I really can not figure out wat u mean in last reply. Like, if 4.31 were to be the -log of H, then wat do we mean by the pH of 4.69?

H⁺ has both concentration and activity. 4.31 is -log(activity), 4.69 is -log(concentration). pH is defined as the latter. Beware - these were calculated differently! See explanation below.

Also I calculated the ph = 4.69 by averaging the two Pkas, also suceeded in computing the values of activity coefficient of by using Dybye-Hukel Equation in calculation center . This value hold with value computed by BATE. However, if i use activity co-ef, I cant get the value of 4.45 by using 4.69 as pH. Instead, the pH increases value increase above 4.69

4.69 is a correct result obtained if all calculations are done using concentrations. However, if you want to use activity coefficients situation complicates. I have addressed it in my previous post. You must do the calculation interatively - few steps (perhaps even 1) will be enough and you will get the 4.31/4.44 results. As activity coefficients are always less then one activities (called eons ago "effective concentrations") are always smaller than the concentrations - thus dissociation goes a little bit further if calculated with the activities. That's why the real pH is lower then 4.69.

In last reply u mentioned some numerical method for obtaining. Could you please help me find that method. or provided me some reference for the same

No idea where to look for it (but there must be plenty of internet sources). Google for Newton method of polynomials solving. BATE solves hard coded polynomial describing the solution. See pH calculation by BATE to find out the equation used (it is converted to the polynomial in the program). But it is not suitable for hand calculations.

Also in the formula for diprotic acid, ( pH = 1/2(pK1 + pK2)) . I cant figure out the concentration of H2A and A- are assumed to be same

Please reword your question.