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# Why does the red blood cells don’t have possess a nucleus ???

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 is n’t right because it should be 4 so I ca n’t trust the standard buff solutions ! ! ! I need to do some experiments which require that I accurately calibrate the ph meter which means I ca n’t titrate the standard buffers with HCl or NaOH. I have to know the exact quantities. Might seem a little extraordinary, but I want to be in truth certain about the mathematics.

I intend to prepare my own calibration standards using claim quantities of chemicals. full concentration should be 0.1 M, volume needs to be 100 milliliter. I think 0.1 M is a good choice because that should have broken ionic lastingness and not a lot crazy debye-huckel variations in ph. I am reasonably sealed that I know how to prepare a ph = 4 standard buffer solution using acetic acerb and sodium acetate. But I am not quite certain how to prepare a ph 7 standard. First of all, is this precisely how I should prepare a 0.1 M acetate fender in 100 mL ? According to pubchem the pKa of acetic acid is 4.76. I have a chem casebook which tells me that the pKa is 4.74. not much difference so I will assume pubchem is veracious. HA + H2O — > A- + H3O+ acetic Acid + H2O — > Acetate + H3O+ Henderson-Hasselbalch equality : ph = pKa + log ( [ A- ] / [ HA ] ) ph = pKa + log ( [ Acetate ] / [ Acetic Acid ] ) = 4.76 + log ( [ Acetate ] / [ Acetic Acid ] ) For the entire concentration : [ Acetic Acid ] + [ Acetate ] = 0.1 M The coincident 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 … reasonably much precisely 4 ! I have arctic acetic acid and sodium acetate rayon. Glacial Acetic acid is a arrant liquid so its concentration is its density / ( Formula Weight ) [ Glacial Acetic Acid ] = [ ( 1.05 g/mL ) / ( 60.05 g/mole ) ] * ( 1000 milliliter / 1 L ) = 17.5 M basic Dilution equation : initial book = ( 0.0852 M / 17.5 M ) * 100 milliliter = 0.487 milliliter arctic acetic acid in 100 milliliter entire bulk. sodium Acetate — > Na+ ( aq ) + Acetate- ( aq )

grams sodium acetate rayon = ( 0.0148 moles sodium acetate / L ) * ( 82.03 grams sodium acetate / 1 mole sodium acetate ) * ( 1L / 1000 milliliter ) * 100 milliliter = 0.121 grams sodium acetate So I can precisely prepare a ph = 4 is 0.1 M acetate solution by adding 487 microliters of glacial acetic acidic & 0.121 grams sodium acetate to 100 mL volumetric flask, and bring the volume up to 100 milliliter with ultrapure deionized water ? Swirl to mix. — — — — — — — — — — — – I am confused about how to prepare an precisely ph = 7 phosphate buff solution. I know the tone wise dissociation of Phosphoric Acid. But the trouble is that I am not certain about the second gear dissociation ceaseless. H3PO4 — > H2PO4- ( aq ) + H+ ( aq ) H2PO4- ( aq ) — > HPO42- ( aq ) + H+ ( aq ) If I know the second dissociation constant of phosphorous acid I could figure out precisely how to make a ph = 7 standard 0.1 M phosphate buffer solution using phosphate salts. My undergraduate general chem textbook book is telling me the second gear 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 claim quantities. But I am not certain about what the second pKa of phosphorous acid is ! Are undergraduate chem textbooks equitable wrong ? That ‘s besides much of a allowance of error. ph = pKa + log [ ( K2HPO4 ) / ( KH2PO4 ) ] besides 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 solid acid HCl with a strong basis NaOH. Therefore the equivalence point of the titration of a strong acerb and potent base should yield a ph 7 of neutrality forming NaCl. so if I dissolved NaCl in body of water to yield a 0.1 M concentration would the ph be equitable angstrom good as making that phosphate buffer solution ? I besides need to figure out how to make a ph = 10 standard buffer solution. Any advice ? Can you make my life easier ? Thanks so much !

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Category : Tech 