So I have some pH = {4, 7, 10} standard buffer solutions from Fisher Scientific. But they must have went bad or I must have contaminated them because when I try to calibrate my pH meter to 4 using the pH 4 standard buffer solution, it calibrates the pH meter to 2 which obviously isn't right because it should be 4 so I can't trust the standard buffer solutions!!!
I need to do some experiments which require that I accurately calibrate the pH meter which means I can't titrate the standard buffers with HCl or NaOH. I have to know the exact quantities. Might seem a little over-the-top, but I want to be really certain about the math.
I intend to prepare my own calibration standards using exact quantities of chemicals. Total concentration should be 0.1 M, volume needs to be 100 mL. I think 0.1 M is a good choice because that should have low ionic strength and not much crazy debye-huckel variations in pH.
I am pretty certain that I know how to prepare a pH = 4 standard buffer solution using acetic acid and sodium acetate. But I am not quite sure how to prepare a pH 7 standard.
First of all, is this exactly how I should prepare a 0.1 M acetate buffer in 100 mL?
According to pubchem the pKa of acetic acid is 4.76. I have a chem textbook which tells me that the pKa is 4.74. Not much difference so I will assume pubchem is right.
HA + H2O --> A- + H3O+
Acetic Acid + H2O --> Acetate + H3O+
Henderson-Hasselbalch Equation: pH = pKa + log([A-] / [HA])
pH = pKa + log([Acetate] / [Acetic Acid])
= 4.76 + log([Acetate] / [Acetic Acid])
For the total concentration:
[Acetic Acid] + [Acetate] = 0.1 M
The simultaneous solution to these equations is about:
[Acetic Acid] = 0.0852 M
[Acetate] = 0.0148 M
pH = 4.76 + log(0.0148 / 0.0852) = 3.9998 etc ... Pretty much exactly 4!
I have glacial acetic acid and sodium acetate.
Glacial Acetic acid is a pure liquid so its concentration is its density / (Formula Weight)
[Glacial Acetic Acid] = [(1.05 g/mL) / (60.05 g/mole)] * (1000 mL / 1 L) = 17.5 M
Basic Dilution Equation:
Initial Volume = (0.0852 M / 17.5 M) * 100 mL = 0.487 mL glacial acetic acid in 100 mL total volume.
Sodium Acetate --> Na+ (aq) + Acetate- (aq)
grams sodium acetate = (0.0148 moles sodium acetate / L) * (82.03 grams sodium acetate / 1 mole sodium acetate) * (1L / 1000 mL) * 100 mL = 0.121 grams sodium acetate
So I can exactly prepare a pH = 4 is 0.1 M acetate solution by adding 487 microliters of glacial acetic acid & 0.121 grams sodium acetate to 100 mL volumetric flask, and bring the volume up to 100 mL with ultrapure deionized water? Swirl to mix.
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I am confused about how to prepare an exactly pH = 7 phosphate buffer solution.
I know the step wise dissociation of Phosphoric Acid. But the problem is that I am not certain about the second dissociation constant.
H3PO4 --> H2PO4- (aq) + H+ (aq)
H2PO4- (aq) --> HPO42- (aq) + H+ (aq)
If I know the second dissociation constant of phosphoric acid I could figure out exactly how to make a pH = 7 standard 0.1 M phosphate buffer solution using phosphate salts.
My undergrad general chem text book is telling me the second Ka is 6.2 * 10^(-8) or the pKa is 7.208 and pubchem is telling me the second the second pKa is 7.09.
Thus I could use the Henderson-Hasselbalch Equation again with the total concentration to figure out the exact quantities. But I am not certain about what the second pKa of phosphoric acid is! Are undergrad chem textbooks just wrong? That's too much of a margin of error.
pH = pKa + log[(K2HPO4) / (KH2PO4)]
Also I was wondering if a 0.1 M solution of NaCl would have a pH of EXACTLY 7? Would it be more or less accurate than making the phosphate buffer?
NaCl is formed by the titration of a strong acid HCl with a strong base NaOH. Therefore the equivalence point of the titration of a strong acid and strong base should yield a pH 7 of neutrality forming NaCl.
So if I dissolved NaCl in water to yield a 0.1 M concentration would the pH be just as good as making that phosphate buffer solution?
I also need to figure out how to make a pH = 10 standard buffer solution.
Any advice? Can you make my life easier?
Thanks so much!