(Ⅰ) According to the problem statement, the trajectory of point $M$ forms a parabola with focus $F$ located at the origin. Thus, the equation of curve $C$ is
$$y^2 = 4x.$$
(Ⅱ) Let's consider the line $l: y = kx + m$. To find the tangent point with the curve $C$, we solve the system of equations:
$$\begin{cases} y^2 = 4x,\\ y = kx + m. \end{cases}$$
Substituting $y$ from the second equation into the first, we get:
$$(kx + m)^2 = 4x.$$
Expanding and rearranging terms, we have the quadratic equation
$$k^2x^2 + (2mk - 4)x + m^2 = 0.$$
To be tangent at a single point, the discriminant $\Delta$ of this quadratic equation must be zero:
$$(2mk - 4)^2 - 4m^2k^2 = 0$$
$$\Delta = 16(1 - mk) = 0.$$
From this, we deduce that $m = \frac{1}{k}$. Substituting $m$, the equation of line $l$ becomes
$$y = kx + \frac{1}{k},$$
and the $y$-intercept $P$ is at $(0, \frac{1}{k})$.

The quadratic equation now becomes
$$k^2x^2 - 2x + \frac{1}{k^2} = 0,$$
solving for $x$ yields $x = \frac{1}{k^2}$.
Subsequently, we find $y = kx + \frac{1}{k} = \frac{2}{k}$, which gives us the coordinates of point $A$ as $\left( \frac{1}{k^2}, \frac{2}{k} \right)$.

Since $Q$ is the midpoint of $OA$, its coordinates are
$$Q\left( \frac{1}{2k^2}, \frac{1}{k} \right).$$

Finally, the slope of $PQ$ is derived from the change in $y$ over the change in $x$:
$$k_{PQ} = \frac{\frac{1}{k} - \frac{1}{k}}{\frac{1}{2k^2} - 0} = \frac{0}{\frac{1}{2k^2}} = 0.$$

Therefore, the slope of the line $PQ$ is
$$\boxed{0},$$
which means the line $PQ$ is horizontal.