Yes, you can have a particle with very small momentum and a very uncertain position. A simple case in which you can work everything out analytically is the wave function
##\psi(x) = e^{-x^2/L^2}##
If you haven't done this before you should calculate various quantities for this wave function like ##\langle x \rangle##, ##\langle p \rangle##, ##\Delta x = \sqrt{\langle x^2 \rangle - {\langle x \rangle}^2}##, and ##\Delta p = \sqrt{\langle p^2 \rangle - {\langle p \rangle}^2}##. It's also enlightening to calculate the time evolution of this wave function, which isn't too hard. If you do end up doing this, make sure to do it again with the more general wave function ##\psi(x) = e^{-x^2/L^2}e^{i k x}##
You'll find that the uncertainty in position is of order ##L##. The expectation of momentum is 0, with uncertainty of order ##\hbar/L##. Taking ##L## large gives a high position uncertainty and a small momentum uncertainty.
This is not a problem with quantum mechanics; it's a straightforward consequence of it. In fact, the intuitive content of the uncertainty principle can basically be summarized as: "If you want a particle to have a certain momentum (such as zero) with high precision, then its position must be very uncertain."