The intensity of a sound is a measure of the energy that the sound waves carries through space per unit time per unit area (since energy is conserved along the wave unless work is done on/by it and the wave spreads out through space).
$$I = \frac{\text{power}}{\text{area}}$$ 
If the wave propagates with amplitude $A$ at a speed $v$ (or frequency $f$ and wavelength $\lambda$) through a medium of volumetric mass density $\rho$, its intensity is (supposedly) given by $$I = {A^2 \over 2\rho v} = {A^2 \over 2\rho f\lambda}$$
In the commercial world, however, we describe sound with sound intensity level in decibels $\rm dB$ of an airborne sound wave of sound pressure (amplitude) $A$ given by
$$L_A= 10\log_{10}{A^2 \over A_\text{ref}^2} \ \rm dB$$
where $A_\text{ref}= 20\ \rm µPa$ is the most common reference pressure, corresponding to the minimum intensity for the average person to hear a sound—if the wave oscillates at $1\ \rm kHz$.
I’m left with one observation and hence one question: the definitions of intensities given are purely mechanical—they don’t take into account subject measurements (collected, for example, by the ISO) of what sounds seem louder than others. A flaw with quoting intensities in watts per square metre or decibels (as I see it) is that a change in intensity in $\rm W/m^2$ or $\rm dB$ is not systematically related to perceived differences by definition.
So what gives? What am I missing?
If you’re still confused, think about radiometry and photometry: irradiance is given in $\rm W/m^2$ as well, but it is scaled by a variable luminosity function to give the illuminance in lux $\rm lx$ that humans sense. The defining relationship is
$$\Phi_{\rm v} = \int\limits_{300\ \rm nm}^{800\ \rm nm} K_{\rm cd} \, V(\lambda) \, \Phi_{\rm e}(\lambda) \, d\lambda$$
