31 TAOLE 26 some weak acids and their conjugate bases Acid (Proton Donor) Conjug

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31 TAOLE 26 some weak acids and their conjugate bases Acid (Proton Donor) Conjugate Base (Proton Acceptor) 1.78 x 10 4,76 1.74 x 10 3.86 1.38 x 10 2.14 7.24 x 10 a Dihydrogen phosphate ion 6,86 Diydrogen phosphate ion 1.38 x 10 Monohydrogen phosphate ion 12.4 3.98 x 10 Monohydrogen phosphate ion 6.3 5.1 x 10 Bicarbonate ion +H+ 10.25 5.62 x 10 989 1.29 x 10 9.25 5.62 x 10 +H Appurant and K. values (see text for explanation pr. This is really an oversimplification because the H,o+ isitself solvated equal. The activities of pure liquids and solids have a value ofone, and by other water molecules, and protons in aqueous solution are trans- in biochemical reactions the activity of the solvent water is often as- erred from one water molecule to another with a frequency of about sumed to have an activity of one. Thus, an alternative expression for per second. the equilibrium shown in Equation 2.5 is: For the purposes ofthe following discussion, it suffices to describe ionization process in an even simpler way, he H+ OH (2.5) To correctly evaluate Equation 2.7, we must use unitless values for long as we remember that a proton never exists in aqueous solu- and OH] that have the same magnitude as their concentra- as a free ion-it is always associated with one or more water mol tions expressed in units of molariry (moles per liter). Throughout this ules. Whenever we write a reaction involving aqueous H we are text we will adopt the convention of indicating molar concentration ly referring to a hydrated for with square brackets around the symbol for that The described by Equation 2.5 can be expressed in Because Ke is a constant, [H+1and oHl vary indepen- ms product which is 10 C: dently If we change either IH or OH) by adding acidic or basic

Explanation / Answer

Make a buffer whose pH is 7.0 by choosing from the acids shown in the table below, and explain your choice.

For a pH near 7... we must choose a pKa vlaue near 7.

So... when using the Henderson HAseelbach equation for buffers:

pH = pKa + log(conjugate base / Acid)

we maximize the log(ratio) section

If we want a strong ratio, then ratio = 1, so there is equal amount of acid and base

so

pH = 7 then

pH = pKa + log(ratio)

pH = pKa approx for max

so

from the list, the nearest vlaue will be 6.86; which is dihydrogen phosphate + hydrogenphosphate

so...

NaH2PO4 + Na2HPO4 will be a good buffer for this

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