What is the pH of 1M Glycerol?
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 $\ce{C3H7O2OH + H2O &lt;=&gt; C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O &lt;=&gt; C3H7O2OH2^+ + OH^-}$, 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 $pH = 6.88$.