"Contradiction" about Hubble parameter $H_0$ in the future? The Hubble parameter $H_0$ is expected to level off asymptotically in the far future as dark energy becomes dominant over matter, radiation, and space curvature. I think it's predicted to go down to 55 km/s/Mpc.
Yet dark energy is also the cause of a galloping acceleration of distant galaxies in the visible universe. Doesn't that imply the straight-line giving H0 today of 67-70 on a Hubble expansion graph will bend upwards to make a greater slope. Doesn't that mean H in the far future will increase dramatically?
I teach this stuff to adult amateur astronomers and still can't make sense of this "contradiction".
 A: The best way to understand what the Hubble parameter is going to do is to look at the behavior of the Friedman scale factor $a(t)$... in other words, just how the universe has been expanding, and will expand in the future.
According to the current standard model of cosmology, the Lambda-CDM model, a good approximation for the scale factor after the first 10 million years is
$$a(t)=(\Omega_m/\Omega_\Lambda)^{1/3}\sinh^{2/3}(t/t_\Lambda)$$
where
$$t_\Lambda=\frac{2}{3H_0\sqrt{\Omega_\Lambda}}.$$
Here $\Omega_m=0.3089$ is the fraction of the energy density due to matter and $\Omega_\Lambda=0.6911$ is the fraction due to dark energy. (This solution of the Friedmann equations neglects radiation, but the effects of radiation on the expansion became negligible by 100 million years after the Big Bang.)
$H_0$ is the current value of Hubble parameter, $67.7$ kilometers per second per megaparsec or $0.0693$ per gigayear.
These three numerical parameters for the model come from precise observations of the cosmic microwave background.
The evolution of the Hubble parameters is given by 
$$H(t)=\frac{\dot{a}(t)}{a(t)}.$$
One finds that the Hubble parameter, currently $0.0693$ per gigayear, slowly decreases in the future to $0.0576$ per gigayear or $56.3$ kilometers per second per megaparsec.
Graphing these quantities, the scale factor looks like

and the Hubble parameter looks like
.
The horizonal axis in both graphs is in gigayears. The vertical axis in the first graph is dimensionless because the scale factor is dimensionless. The model normalizes to be $1$ right now, $13.8$ gigayears after the Big Bang. The vertical axis in the second graph is in inverse gigayears.
If by “Hubble expansion graph” you mean a plot of velocity vs. distance, the slope of that linear (near us, at least) graph is going to slightly decrease.
As for the “contradiction” that bothers you, Wikipedia has this to say:

Another common source of confusion is that the accelerating universe does not imply that the Hubble parameter is actually increasing with time; since $H(t)\equiv\dot{a}(t)/a(t)$, in most accelerating models $a$ increases relatively faster than $\dot a$, so $H$ decreases with time. (The recession velocity of one chosen galaxy does increase, but different galaxies passing a sphere of fixed radius cross the sphere more slowly at later times.)

A: The Hubble parameter does not measure the rate of change of expansion. It is the ratio of the expansion rate to the size of the universe. $H(t) = v(t)/d(t)$.
If the expansion is accelerating, then the size of the universe gets bigger faster than the expansion rate gets bigger - hence the Hubble parameter gets smaller.
e.g. If you jump off a cliff, what happens to the ratio of your velocity, to the distance you have fallen?
$$ v = gt, \ \ \ \ \ \ d = gt^2/2$$
and hence $v/d$ gets smaller as $1/t$, and ignoring pesky things like the ground and terminal velocity, would head towards zero.
The accelerating universe behaves slightly differently in that $v$ tends towards an exponential with time as the vacuum energy becomes dominant:
$$ v \rightarrow \exp( \alpha t)$$
Hence
$$ d \rightarrow \left(\frac{1}{\alpha} \right) \exp(\alpha t)$$
and
$$ H(t) = \frac{v}{d} \rightarrow \alpha$$
The net behaviour is similar. $v/d$ starts out big, but trends asymptotically to a fixed value as the velocity of expansion accelerates exponentially. That fixed value turns out to be $\sqrt{\Lambda/3}$.
