Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: BHAVESH on March 25, 2020, 05:12:44 AM
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Hi,
Is there any way to calculate pH of water based on Ionic Strength?
Thanks
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Yes, there is.
Check your textbook, Wikipedia (Self-ionization of water) and Chembuddy's pH calculation lectures
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I mean natural water bodies having different ion concentration. So if analyze these ions, can we calculate pH using ionic strength of that particular water body?
And Can Van't Hoff Equation be applied here as well???
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It is enough to calculate the ionic strength, H3O+ ion activity coefficient and apply a precise definition of pH. Condition - water must not contain hydrolyzing ions.
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Assuming pH is a function of two things only - water autoionization and activity coefficients that depend on the ionic strength - yes, it is enough to knwo concentration of all ions present.
Trick is, with natural waters these assumptions will be never correct, as they contain plenty of acids/bases that change the pH. Almost every ion present, be it cation or anion, reacts with water slightly changing pH. In the case of ions like Na+ or Cl- effects can be safely ignored, in the case of ions like Ca2+ or CO32- - present in every natural water - the effect is quite prominent.
It is actually much easier to just measure pH than to do full analysis of all ions present. Please remember what we measure is not the concentration, but activity of ions, so the pH measured is already corrected for the ionic strength.
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I fully agree on what you say. Now for an example, in Reverse Osmosis systems; feed pH is measured and then pH in reject & permeate is calculated based on alkalinity and pH of feed water using following equation.
CO2f = Alkf X exp-((pHf-6.3022)/0.423)
then, pH of Concentrate(Reject) is calculated using alkalinity/CO2 ration of concentrate water with following equation.
pHc = 0.423 X Ln(Alk/CO2ratio) X 6.2033
The reference of above equations are given in ASTM - D4582 for the calculation.
Now my question is, in Reverse Osmosis; since membrane is semipermeable CO2passes through membrane and thus we get lower pH of permeate water. When we calculate pH of concentrate water, it is on higher side than the actual pH measured. How??
Secondly, I want to co-relate carbonate iion conversion from bicarbonate ion at specific pH of concentrate at constant given temperature.
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And I have already calculated Activity Coefficient using Davies equation which is
Log[γz] = -A X Z2[√I/(1+ √I)]-0.3I
Where, A = 1.82 X 106 X [ET] - 1.5
I = Ionic Strength
E = [60954/(T+116)] - 68.937
T = Temp in °K
Now I am unable to catch, where am I missing?
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Looks like you are trying to calculate pH ignoring all acid/base equilibria present. No way. You have to start with these equilibria, ionic strength is the last factor to be taken into account.
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No Borek, I have already calculated Temperature Coefficients for a Variety of Acid-Base and Solubility Constants. Not only this, I have calculated pK, K, K' & pK' of various Acid-Base and calculation of saturation pH.
I used following equation for Acid-Base to calculate pK.
Formula used for pK's: pK = a1 + a2/T + a3Log[T] + a4T + a5/T2
a1, a1, a3 a4, & a5 figures have been referred from various books like
Nordstrom etal, 1990
Data in Sillen & Martell, 1964,1971
Harned & Owen, 1958
Plummer & Busenberg, 1982
etc...
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All the time you talk about calculating constants, not about finding the equilibrium they describe.
Let's start with something simple, to make sure we are talking about the same thing. If I tell you I did analysis of a solution and I know it contains 0.03 M Na+, 0.01 M HCO3- and 0.01 M CO32-, will you be able to tell me what is pH of this solution?
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What I understood from your question that it is a solution of 0.01 M Na2CO3 & 0.01 M of NaHCO3. But its not clear how shall I calculate pH.
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What I understood from your question that it is a solution of 0.01 M Na2CO3 & 0.01 M of NaHCO3. But its not clear how shall I calculate pH.
You see, it is a basic, HS level problem, one that can be solved with a reasonable accuracy just by looking at it and checking pKa values for carbonic acid. It doesn't require any fancy formulas for pK values, nor ionic strength of the solution.
I strongly suggest you read any general chemistry textbook, section on the equlibria calcuation, especially acid base equilibria and pH calculation. Plenty of free texts on the web. At the moment you are trying to run without knowing how to walk.
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Thanks Borek for your advise.
pH will be 12.0
I used Ka · Kb = 1 X 10-14
Please correct me if I am wrong.
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You are wrong. pH should be close to 10.
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pH is going to be 10.32
Using Henderson-Hassel Equation using Ka2 = 4.8 X 10-11
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And now correct this value with ionic strenght of solution.
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can you please give some hint??
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Calculate activity coefficients.
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Shall I use Davies Equation?
Log[γz ]= -A* Z2 [√I/(1+ √I)]-0.3I
A = 1.82X106*[ET]-1.5
E = 60954/[T + 116] - 68.937
T = Temp in °K
I = Ionic Strength
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For the beginning use standard temperature 25°C.
Check your Davies equation for the correct syntax.
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Thanks AWK
Log[γz]= -A*Z2[√I/(1+ √I)-0.3I]
I hope this is correct now. Thanks :D :D :D
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I Calculated Activity Coefficients
Ionic Strength - 0.2499
Temp. - 25°C = 298.15°K
A Value - 0.511
E Value - 78.24
Ionic Charge γ Value
1 0.738
2 0.297
3 0.065
4 0.008
Now, what should be the next step?
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For the problem given by Borek, the ionic strength is close to 0.04.
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Ok
Ionic Strength is 0.04 using I = 0.5ΣCiZi2
Temp. - 25°C = 298.15°K
A Value - 0.511
E Value - 78.24
For,
Ionic Charge γ Value
1 0.834
2 0.483
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Now calculate pH of this solution with activity coefficients and compare with the value obtained without activities.
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Ok
Ionic Strength is 0.04 using I = 0.5ΣCiZi2
Temp. - 25°C = 298.15°K
A Value - 0.511
E Value - 78.24
For,
Ionic Charge γ Value
1 0.834
2 0.483
Division of γ Value = 0.834/0.483 = 1.7267
K - 4.8 X 10-11
K' =4.8 X 10-11 X 1.7267 = 8.2882 X 10-11
Corrected pH = -Log[K'] = 10.082
Original pH is 10.32
Please correct me if I am wrong.
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And where is the activity of H+ ions?
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Ok
Ionic Strength is 0.04 using I = 0.5ΣCiZi2
Temp. - 25°C = 298.15°K
A Value - 0.511
E Value - 78.24
For,
Ionic Charge γ Value
1 0.834
2 0.483
Division of γ Value = (0.834 X 0.483)/0.483 = 0.834 (Product/Reactant)
K - 4.8 X 10-11
K' =4.8 X 10-11 X 0.834 = 4.0032 X 10-11
Corrected pH = -Log[K'] = 10.40
Original pH is 10.32
Does it mean that pH always increases when we use Ionic Strength as a correction factor, OR does it depend upon what kind of Acid, Base or salt reaction we work with???
Please correct me if I am wrong.
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The value of K' is ambiguous. What exactly does it mean for you? Calculations can be made using the thermodynamic constant K. I expect a pH value very close to 10.
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What I think is that at constant temperature, ionic strength affect K value OR pH Value. When It is corrected using ionic strength, gives us actual value.
Can you please give me some hint on using thermodynamic constant K?
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K=aH3O+ · aCO32- / aHCO3-
aH3O+ = K · aHCO3- / aCO32-
pH = pKa + log(aCO32-/aHCO3-)
a= γ · c
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pH will be 9.844
Ok the mistake was I was doing that
1) I was not considering given concentration of HCO3- & CO3-2
2) In back of my mind, I was considering H3O+ ion concentration also as 0.01 M
3) I was not considering Equation of K = [CO3-2][H3O+]/[HCO3-] in which Standard K is already given as 4.8 X 10-11
4) I was not considering concentrations of all ions with correction of γ Value.
5) I did not have in my mind to use Handerson equation there after.
Thank you very much AWK for teaching me.
IF any of my mistake is left apart from above mentioned, please do let me know.
Wonderful teaching..