What is the pH of 1M Glycerol?

Question

If the pKa of Glycerol is 14.15. How do you calculate the pH for it?

I assume that the Henderson-Hasselbalch derivative that works for weak acids and bases is not applicable here.

Answer 1

The answer is approximately 6.88. Therefore a $1\ M$ solution of glycerol in water will be ever so slightly acidic (considering that the hydroxyls are much weaker bases than acids, i.e. that the equilibrium constant for the reaction $\ce{C3H7O2OH + H2O <=> C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O <=> C3H7O2OH2^+ + OH^-}$, which may or may not be true. Inclusion of the second equilibrium will drive the true pH even closer to 7)

For very weak or very dilute acids/bases, solving a dissociation problem is a bit more tricky because the self-dissociation equilibrium of water must be taken into account. It is often a valid approximation to forget about it, as water has only a small tendency to self-dissociate and so its effect on equilibrium concentrations of $\ce{H^+_{(aq)}}$ and $\ce{OH^{-}_{(aq)}}$ is often negligible, but this is not your case; glycerol in water seems to be about as weak an acid as water itself.

The best way to solve equilibrium problems is to set up a large equation including all factors by combining smaller equations. We need to make an equation that takes into account all the species in the medium ($\ce{H^+_{(aq)}}$, $\ce{OH^{-}_{(aq)}}$, $\ce{C3H7O2O^{-}_{(aq)}}$ and $\ce{C3H7O2OH_{(aq)}}$ {for convenience I shall label the latter two $\ce{A^-}$ and $\ce{HA}$}) and the related equilibrium constants ($K_a$ and $k_w$). The smaller equations are found by balancing charges (one equation) and the amount of matter of each substance (one or more equations depending on the problem).

Charge balance: $[H^+]=[OH^-]+[A^-]$ (I)

Matter balance: $[HA]+[A^-]=C_{acid}=1\ M$ (II)

The equations for the equilibrium constants are:

$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-14.15}$ (III)

$k_w=[H^+][OH^-]=1\times 10^{-14}$ (IV)

To solve the problem, we need to play with the equalities until we get a single equation with a single variable. It's convenient to obtain an equation in $[H^+]$ because we are able to find the pH directly by applying $-log$ to the answer. Notice that

$[OH^-]=\frac{k_w}{[H^+]}$ (V)

$[HA]=\frac{[H^+][A^-]}{K_a}$ (VI)

Substituting (VI) in (II) yields, after some algebra:

$[A^-]=\frac{C_{acid}K_a}{[H^+]+K_a}$ (VII)

Now insert (V), (VI) and (VII) in (I) so that you get:

$[H^+]=\frac{k_w}{[H^+]}+\frac{C_{acid}K_a}{[H^+]+K_a}$

A little bit of persistence will get you to the following polynomial:

$[H^+]^3+K_a[H^+]^2-(C_{acid}K_a+k_w)[H^+]-K_ak_w=0$

Replacing the values of $K_a$, $k_w$ and $C_{acid}$, the only positive root for the equation is $[H^+]=1.30688\times 10^{-7}$, which after applying the antilogarithm results in $pH = 6.88$.

Answer 2

Correct, Henderson-Hasselbalch is for when you have a buffered solution of both an acid and its conjugate base.

Here you want to imagine dissolving one mole of Glycerol in one liter of water. When we say that the pKa of glycerol is 14.15 we mean that the Ka for the reaction:

$$C_3H_7O_2OH \Rightarrow C_3H_7O_2O^- + H^+$$

is $10^{-14.15}$. This is a very small number, which indicates that while glycerol dissolved in water does produce some $H^+$, it likely won't decrease the pH very much. So here you would expect a pH slightly less than 7.

To calculate this value, I'd construct what many call an "ICE chart" Example 2 in this document seems similar to the problem you're working on.

Answer 3

I know a formula relating $K_a$,dissociation constant A and concentration C for weak acids and bases as $K_a$=$A^2$C.This may help you out.