CMOS
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calculating pH of salts of polyprotonic acids
Hi
I have some problems with my lab report. We were measuring pH of salts, and since the pH-meters were giving some weird values(e.g. 5.6 for freshly
distilled H2O) we were asked to calculate theoretical values. I calculated pH for salts of monoprotonic acids(and bases like
Zn(OH)2 and Fe(OH)3) but I made mistakes in calculating for
NaH2PO4,Na2HPO4,KAl(SO4)2 and K2SO4(in these two it's probably
because I didn't considered SO42- dissociation). I did some research and find this. This method(p. 26th) gave me different results for Na2HPO4 and Na2CO3 but I got the same number
for NaH2PO4, pH=7.06 for C=0.01 . Calculating for first time, I was using normal quadratic equation (x2+x*K-K=0),
Kw=Ka*Kb and K=K1*K2*K3(and here was my mistake because I was using this K for
solving equation). I couldn't find similar subject so I'm asking here. English isn't my first language so sorry for any mistakes, and thanks for any
help.
[Edited on 17-4-2014 by CMOS]
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Tsjerk
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You can't really measure the pH of dH2O, because there are to little free protons (or something like that), just like the pH of the KCl solution you
use to keep the electrode in is not ever given as seven.
The values I get from the KCl solution sometimes differ 2 or 3 pH points after a measurement compared to before the measurement, but when measuring
something that actually contain an acid or base, it works fine.
[Edited on 17-4-2014 by Tsjerk]
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blogfast25
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These values can only realistically be estimated by simplified theory.
For NaH2PO4, just take the Ka2 value of H3PO4 and work out the quadratic equation.
For Na2HPO4, consider this a weak base and work out the Kb from Ka3. Use the simple version of pH calculation for a weak base.
For K2SO4, assume this to be almost neutral: salt from a strong acid and a strong base.
For KAl(SO4)2, assume full dissociation into K+, [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> and sulphate ions.
Only [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> affects pH because it behaves like a weak acid. But finding a
Ka value for this ion may not be easy. Once found, just apply the quadratic equation again.
[Edited on 17-4-2014 by blogfast25]
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DraconicAcid
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Actually, for an amphoteric ion at reasonable concentrations, the pH is pretty close to the average of the pKa of the ion and the pKa of its conjugate
acid.
For NaH2PO4, its pKa is7.2. The pKa of its conjugate acid is 2.15. The average of those is 4.68. Merck Index says its pH is
actually 4.5, which is close enough.
You can derive that by assuming that the main reaction in solution is the amphoteric ion reacting with itself. Find the Keq for that from Ka and Kb
of the ion, and solve to find the ratio of the ion to the acid and the base. This will give you the pH.
ETA: Generic acid is H2A, with K1 and K2.
The reaction of 2 HA- = H2A + A2- will have an equilibrium constant of K2/K1. If we start with
an initial concentration of C for the hydrogen generate ion, x amount of this will react, giving [HA] = C-x and [H2A] = [A2-] =
x.
So K2/K1 = x2/(C-x)2
Take the square root of both sides to get the ratio of x/(C-x).
Now look at the acid ionization of hydrogen generate ion. HA- + H2O = H3O+ + A2-
K2 = [H+][A=]/[HA-]
Agebraically jiggery-poke it around, and [H+] = K2(C-x)/x
Which means that [H+] = K2 times the square root of K1/K2, which works out to be the square root of the product of
K1 and K2.
Take the negative log of both sides, and pH = (pK1 + pK2)/2
[Edited on 17-4-2014 by DraconicAcid]
ETA: A different derivation of the same result is given on page 28 of the OP's link.
[Edited on 17-4-2014 by DraconicAcid]
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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nezza
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Distilled water nearly always gives an acidic pH due to dissolved CO2 which reacts as a weak acid.
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chornedsnorkack
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Quote: Originally posted by CMOS ![](images/xpblue/lastpost.gif) | Hi
I have some problems with my lab report. We were measuring pH of salts, and since the pH-meters were giving some weird values(e.g. 5.6 for freshly
distilled H2O) |
5,6 is the correct value for pure water saturated with the solution of 0,03 % CO2 in air.
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Metacelsus
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Actually, it's 400 ppm these days .
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Mildronate
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You need to write charge and mass bilance and then solving polynomal equation http://www.chem.mun.ca/courseinfo/c3100/IV.%20Acid%20Base.pd...
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