acid equilibria and alpha plots

Acid equilibria and alpha plots

Chemistry 321, Summer 2014

In this lecture

• Alpha plots show dynamic changes in species concentration during a titration

• For weak acids, alpha plots mirror the behavior seen on titration plots and yield further information

Formal concentrationGiven:

Let CHA = formal concentration of the acid= [HA] + [A–] = sum of all forms of the

acid at equilibriumCHA = [HA]0 if only the weak acid is added

= [A–]0 if only the conjugate base is added= [HA]0 + [A–]0 if both are added

Alpha is the mole fraction of a species relative to the solution’s formal concentration

So, symbolically, For instance,

We can rewrite this like: [HA] = αHA CHA or [A–] = = αA– CHA

The constraint is that Σ αi = 1

For a monoprotic acid, αHA + αA– = 1

The equation of the alpha plot

Start with the definitions of CHA and αHA:



Invert the expression:



Invert the expression:

Recalling the equilibrium expression:

so

The equation of the alpha plot (continued)

Re-invert the equation:

Similarly, we can derive an expression for αA–:

Note that both expressions for alpha depend on [H+] (and, by extension, pH) only! So this can be plotted.

-1 1 3 5 7 9 11 13 150

0.10.20.30.40.50.60.70.80.91

Alpha Fractions

HAA-

pH

Alph

a

The alpha plot

Where the two curves cross (each α = 0.5), the x-coordinate is the pKa of the weak acid.

pKa

Limiting behavior on the graphFor acetic acid, Ka = 1.8 × 10–5. Consider an acetic acid solution at pH 8.74 (meaning [H+] = 1.8 × 10–9).

The point is that even at a really high pH, there is still some undissociated acid left (i.e., not zero).

When you are dealing with a polyprotic acid or base, then there are pHs at which some species have zero concentration…this is also important when dealing with ligands.

Polyprotic acid alpha plots

Polyprotic acid dissociation occurs in a stepwise manner; that is, the different H+ ions on each molecule dissociate at different pHs, rather than all at once. Note the definitions of the equilibrium

constants Ka1 and Ka2

N.B.: pKa1 = 6.37, and pKa2 = 10.25

Setting up the alpha plot equations for a diprotic acid

Let CH2A = [H2A] + [HA–] + [A2–] ( = formal concentration

of H2A)

Expanding the previous definition of αi:

[H2A] = αH2A CH2A and [HA–] = αHA– CH2A and [A2–] = αA2- CH2A

Note that αH2A + αHA– + αA2- = 1


Inverting the equation yields:



Recall the definitions of Ka1 and Ka2:



Recall the definitions of Ka1 and Ka2:

multiply to get:

Setting up the alpha plot equations for a diprotic acidSubstitute into the original expression:

Setting up the alpha plot equations for a diprotic acidSubstitute into the original expression:

Put it all over a common denominator:

Limiting behavior: at low pH (acidic), the [H+]2 term dominates the other terms, so αH2A ≈ 1; at high pH (alkaline), [H+]2 0, so αH2A ≈ 0.

The other αi expressions (αHA– and αA2–) are derived similarly.


The alpha plot for carbonic acid

What are the pKas for carbonic acid?


At pH 4, αCO3 2- ≈ 0, according to the graph


not zero, but basically negligible!

At pH 4, the only two species that matter are HCO3– and H2CO3

The titration curve for carbonic acid reflects the stepwise dissociation behavior

Challenge problemConsider the stepwise dissociation of phosphoric acid:

Challenge problem (continued)• Derive the equations to calculate αi for all phosphate-containing species.

• Draw the alpha plot for all phosphate-containing species; make sure the axes are labeled and the pKas are sensible.

• Calculate the mole fraction (αi) for all phosphate-containing species at blood pH (7.40).

• Assume all phosphate-containing species have a soluble sodium salt (i.e., if you need PO4

3–, you will use Na3PO4). Which two salts will you use to create pH 7.40 phosphate buffer?

acid equilibria and alpha plots

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