Yes. a wave created with a certain frequency in the water remains the same even if the medium is changed. The only thing that is important in here is the amplitude of our wave.
When the sound wave from inside of water hit the surface, some of the wave reflect back into the water and a lower amplitude wave is continued in the air. so the energy of the wave must be enough so the wave stay detectable outside of the water.
Frequency is a property of source and will not change even if the medium is changed.
EDIT: since the refraction is requested, I add this here:
The refraction in the sound wave is just like the refraction in any wave behavioral phenomena. and it can be calculated this way:
$$\frac{n_{Air}} {n_{Water}} = \frac{v_1}{v_2} = \frac{n_2}{n_1} $$
in this way you can calculate the placement of receiver.
Power $(E/t)$ created by any sound source is calculated this way:
$P=2\pi^2A^2f^2\mu v$
(Usually wave sources provides you the power of machine, no need to calculate it)
Then you can calculate the Intensity of the wave around the source (In the water) with this equation:
$I=\frac{p}{4\pi r^2}$
when r is the distance from the source.
so you can calculate the amount of energy at the surface, and the angle that it refracts. and remember the point at the mediums border act like a sound source it self. it seems like the point is creating another wave trough the air with another wavelength. and the new wavelength is calculated with the new speed (since the frequency doesn't change. it's the same as the sound source in the water, 30 KHZ)
$\lambda = \frac{v}{f}$