Also, making an amp for an alternating signal like with toneAC would be a challenge. You can't just use two 2N2222 transistors as you also need to alternate the ground as well, which flips between the two. Not that it's impossible, just much more complex than a typical simple transistor circuit. Keep in mind that toneAC is designed to be like a voltage amplifier without needing an actual circuit. In essence, it doubles the voltage, which increases the volume.
Tim
If you are using toneAC for the sound quality and still want an amplifier for it a simple solution will be two NPN transistors  on both connect the emitter to ground, the collectors to the two poles of the speaker (choose a "positive rail" transistor and a "negative" one and connect each's base to its respective pole of the speaker), and the base through a current limiting resistor (see below for value) to D9 (for the transistor on the "negative" side of the speaker) and through a current limiting resistor (see below for value) to D10 (for the "positive" side transistor).
The final touch is to add two pullup resistors on the collectors of both the transistors (the resistance values of these will limit the current sufficiently by the way so you don't need that 100ohm recommended resistor on the positive speaker pole). To determine these pullup values take 1/2 the vcc (the "+" voltage) of the circuit (should be the same as the high voltage of the digital pins  3.3 on some arduinos, 5v on most) (5*1/2=2.5 volts). The current through each resistor can be determined with:
v=ir (ohm's law)
(v/2)=i*r
i=v/(2r)
r is resistance of the pullup (ohms), i is current through the pullup (amps), v is vcc (volts)
So the current through each resistor is v/(2r) r being the resistance of the pullup. Multiply this by two because you have 2 resistors (one on each collector) supplying current to the speaker  now we get v/r for the current passing through the speaker (r being the resistance of each individual pullup  they should be the same (the math gets more complicated if they aren't)
We can then plug this watt's law to determine the power output of this amp:
p=iv (watt's law)
p=v*v/r
p being power (watts)
now if you want to pass 1.5 watts through a speaker (say if its rated for 2 watts) you plug 1.5 in (and lets use 5v as vcc  we'll pretend we are using an arduino uno)
p=v*v/r
1.5=5*5/r
r=25/1.5
r=16.67 ohms
So you should have each resistor be 16.67 ohms  remember the resistors should have an adequate power rating ( at least 1/2 the speaker wattage (>(v*v)/(2*r) each so >0.75 watts each if the speaker wattage is 1.5 watts  you should have some buffer room though)
**IF YOU COULDN'T FOLLOW THAT, USE THIS EQUATION: resistance=positive voltage*positive voltage/desired power through the speaker (for an uno: r=25/desired power)
Now for the value of the resistor between the digital pins on the bases of the transistors (they should both be the same as you might have guessed) referred to as base res in the schematic below. We will first start with the desired "collector current" flowing through the transistor (the current that the transistor is giving to our speaker in plain english); we want:
("collector current" rating of the transistor (Ic max))>i>v/(2r)
i being the most "collector current" that the transistor will let through (Ic) and r being the PULLUP RESISTOR ON THE COLLECTOR (we want the current flowing through the transistors to be able to match that of the pullup resistors (v/(2r); calculated above) so it should be designed to handle more current than the resistors will draw  this current must also be below the specified collector current rating for the transistor  choose your transistor wisely)
when choosing a transistor consider: Ic max (see above) and Max power rating (above v*v/(2r), r being the pullup resistor value)
So now look at your chosen transistor datasheet and find a chart or a graph that specifies "current gain" for a given collector current (current going through the transistor)  the collector current will be the "i" value you chose above (greater than v/(2r) and less than the Ic max rating) and find the "current gain" at that given value. Now plug the current gain and the i value (collector current; Ic) into this equation:
Base current= Ic/current gain
Now find the Emitterbase voltage (Vebo) specified in the datasheet and the required voltage drop across the resistor will be: vcc ("+" voltage of the system; 5v if using an uno)  Vebo
So take that voltage drop and plug it into ohms law to find the resistor value:
v=i*r
(vccVebo)=(Ic/(current gain))*(r)
r=Ic/((vccVebo)*(current gain))
Ic was chosen by you above (the "i" value), the current gain was found in the datasheet with Icm vcc was determined at the beginning of this tutorial, and Vebo was found in your transistor dtasheet as well. Now you have an r value for teh resistors on the bases of your transistors. Power rating of these resistors are equal to:
power = Ic*(vccVebo)/(current gain)  this will usually be negligible though so a 1/4 watt resistor should be fine in general
**IF YOU COULDN'T FOLLOW THAT, A 1/4 watt resistor with a VALUE OF 2.2k ohms on each transistor base should be fine***
Enjoy unlimited volume  REMEMBER SPEAKER POWER RATING, RESISTOR POWER RATING (On bases and on pullups), AND TRANSISTOR POWER RATING AND CURRENT RATING to not break stuff
Here is a schematic of the circuit:
v *NPN bjt transistors (EBC = emitter, base, collector)*
<E>GND
D9<base res><B>
<C><Resistance value chosen with above math>VCC

 negative pole of the speaker ()
GND >>>> LOUD SOUND
^<E> positive pole of the speaker (+)
D10<base res><B> 
<C><Resistance value chosen with above math>VCC
Please correct me if I have made any mistakes in my explanation but I have tested this circuit and it works.