Alkaline glycine buffer molarity Having never prepared a buffer solution before, I am hoping for a little help in understanding the process. 

This method I plan to use calls for a $\pu{0.1M}$ glycine buffer at pH $10.6$. Some internet searching yielded this recipe (very bottom of the page) for making a glycine-$\ce{NaOH}$ buffer at $10.6$ pH, which I assume is what I'm looking for.

Most of it seems to make sense to me. $\pu{0.2M}$ stock solutions of both glycine and $\ce{NaOH}$ would be prepared as:

0.2M NaOH solution = 0.2 * 39.9971 = ~8g NaOH/l
0.2M glycine solution = 0.2 * 75.07 = ~15g glycine/l
(Molar mass: NaOH = 39.9971; glycine = 75.07 g/mol)


And, to yield a buffer at $10.6$ pH, mix:

25ml glycine stock + 22.75ml NaOH stock + deionized water to 100ml


Where I get lost is figuring out the molar concentration of the resulting solution, since there are two different solutes. Would it be just the total number of moles present of both, something like:

(25ml + 22.75ml) * 0.2M / 100ml = 0.0955M ?


It seems pretty close to the $\pu{0.1M}$ concentration called for, but I can't seem to find any confirmation of this.
As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one. 

Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.

How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?
$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$

You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:

Total amount of substance in buffer solution:
$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$

Total volume of buffer solution:
$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$

Molarity of buffer solution:
$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$