The position of an electron is known to within 25 pm…

Step 1. Uncertainty. The bottom is $4\pi m \Delta x = 4\pi \times 9.109 \times 10^{-31} \times 2.5 \times 10^{-11} = 2.86 \times 10^{-40}\ \text{kg m}$. So $\Delta v = \dfrac{6.626 \times 10^{-34}}{2.86 \times 10^{-40}} = 2.3 \times 10^{6}\ \text{m s}^{-1}$.

Step 2. Compare. $\dfrac{2.3 \times 10^{6}}{2.19 \times 10^{6}} \approx 1.1$: the uncertainty is larger than the orbital speed itself.