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Offline xshadow

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calculation of pH: approximated formula
« on: February 04, 2016, 12:40:41 PM »
Hi!!

I have some doubt about the calculation   of pH using the correct formula...
For example per the calculation ot the pH of a mixture of 2 weak acid i have the general formula:

[H+]= [A-]' + [A-]'' +[OH-]
' = acid 1
''= acid 2

Where:
[A-]' =  k_a'/(k_a'+[H+]) * C'
[A-]'' = k_a''/(k_a''+[H+]) * C''
[OH-]= K_w/[H+]

[H+]=  k_a'/(k_a'+[H+]) * C'  + k_a''/(k_a''+[H+]) * C''  +  K_w/[H+]

This is the General expression for the pH...but usually i need to do some semplification...
For example:

[H+] >> [OH-]   i have:
[H+]=  k_a'/(k_a'+[H+]) * C'  + k_a''/(k_a''+[H+]) * C''

And this is only the first possible approximation...I could do others.
1) BUT the problem is:  how can i verify if the approximation i used is  correct??  C

2) What does it mean the" >> " from a numerical point of view in these expressions:
[H+] >> [OH-]
K_a' >> K_a''

How much the two terms must be different from a numerical point of view?? At lesson they said to us 10^2 or more? is correct?

3) and if I find:
pH>pk_a
pk_a' > pk_a''
How much must be  the difference?? 1 or 2 unit?? or less? (0,...)

THANKS very much :)

Offline AWK

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Re: calculation of pH: approximated formula
« Reply #1 on: February 04, 2016, 05:51:13 PM »
Quote
How much the two terms must be different from a numerical point of view?? At lesson they said to us 10^2 or more? is correct?

10^2 gives you erron of 5 % in concentration and ~0.02 pH unit
AWK

Offline xshadow

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Re: calculation of pH: approximated formula
« Reply #2 on: February 05, 2016, 05:07:29 AM »
Quote
How much the two terms must be different from a numerical point of view?? At lesson they said to us 10^2 or more? is correct?

10^2 gives you erron of 5 % in concentration and ~0.02 pH unit

Ok thanks :)
So if i have a condition like this:

pH > pKa  ....in this case how much the difference must be to have an error of  5 % ??

If 10^2 (ka>>H+) gives me an error of 5 % in concentration and  an error of 2% in pH unit to have an error of 5 % in pH unit (where H+ becomes pH and ka becomes pka so i get the expression pH>pka ) i have to do:

log_10 (0,005) = -1,3....so the difference must be 1,3 or more in order to use that approximated formula correctly ??

Thanks a lot.
« Last Edit: February 05, 2016, 06:12:47 AM by xshadow »

Offline AWK

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Re: calculation of pH: approximated formula
« Reply #3 on: February 05, 2016, 08:35:15 AM »
My comments concerned to an approximated formula
[H+] = SQRT(Ka1·c1 + Ka2·c2 + ... +Kw)
and pH ≈ -log([H+])
These approximations are valid in most calculations when ionic strength ≤ 0.01

When Ka1·c1 is at least 20 times higher then other parts of expression within square root.
1/20 =0.05 then [H+]=SQRT(c·1.05) ≈ 1.025·SQRT(c)
and -log(1.025)≈1.011

Last part Kw under square root is needed as correction when pH calculated without Kw is greater than 6.8

Examples:
0.1 M acetic acid and 0.5 M formic acid - may be treated as pure formic acid

0.005 M acetic acid and 0.001 M formic acid - both acids should be taken into account in calculations

5·10-10 M acetic acid and 1.0·10-10 M formic acid - addition of Kw is needed under square root

Check my examples by calculations!
AWK

Offline xshadow

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Re: calculation of pH: approximated formula
« Reply #4 on: February 06, 2016, 07:54:48 AM »
My comments concerned to an approximated formula
[H+] = SQRT(Ka1·c1 + Ka2·c2 + ... +Kw)
and pH ≈ -log([H+])
These approximations are valid in most calculations when ionic strength ≤ 0.01

When Ka1·c1 is at least 20 times higher then other parts of expression within square root.
1/20 =0.05 then [H+]=SQRT(c·1.05) ≈ 1.025·SQRT(c)
and -log(1.025)≈1.011

Last part Kw under square root is needed as correction when pH calculated without Kw is greater than 6.8

Examples:
0.1 M acetic acid and 0.5 M formic acid - may be treated as pure formic acid

0.005 M acetic acid and 0.001 M formic acid - both acids should be taken into account in calculations

5·10-10 M acetic acid and 1.0·10-10 M formic acid - addition of Kw is needed under square root

Check my examples by calculations!

So  Ka1·C1 at least 20 times(for a 0,05 error) bigger than  Ka2·C2  is equal to say that:

 Ka1·C1 >>   Ka2·C2 
where  >> is at least 102 (this "gap" give me in fact an error of 0,05)

Is it correct??
Thanks :)

Offline AWK

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Re: calculation of pH: approximated formula
« Reply #5 on: February 06, 2016, 08:11:26 AM »
102 gives you error less than 1% for concentration and for pH less than 0,005 unit when ionic strength is negligible.
AWK

Offline xshadow

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Re: calculation of pH: approximated formula
« Reply #6 on: February 06, 2016, 04:00:44 PM »
But why before you said:

"10^2 gives you erron of 5 % in concentration and ~0.02 pH unit"??

Usually  don't 10^2 give a concentration error of 5%?

Thanks.

Offline AWK

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Re: calculation of pH: approximated formula
« Reply #7 on: February 06, 2016, 04:18:05 PM »
Students better remember 100 then 20. Moreover using 100 their results will be sufficiently correct (within 5 %) even with 1 significant digit.
AWK

Offline xshadow

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Re: calculation of pH: approximated formula
« Reply #8 on: February 12, 2016, 08:10:40 PM »
Students better remember 100 then 20. Moreover using 100 their results will be sufficiently correct (within 5 %) even with 1 significant digit.

IF I have understood correctly for an error of 5% is enough a difference of 20 times,right??

Thanks very much!

Offline AWK

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Re: calculation of pH: approximated formula
« Reply #9 on: February 13, 2016, 02:40:44 AM »
Yes, 20 is sufficient for 5% error in concnetration of H+.
AWK

Offline xshadow

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Re: calculation of pH: approximated formula
« Reply #10 on: February 13, 2016, 01:20:07 PM »
Yes, 20 is sufficient for 5% error in concnetration of H+.

Thanks very MUCH :)

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