# 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: $${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$

• The first answer might be wrong. Please verify, because when you use NaOH to convert glycine to its sodium salt you should actually substract but not add NaOH moles in the total mole calculation and then add them together. As such the molarity will not change. Just think about it the pKa for glycine-COOH is 9.6 and your desired pH is 10.6. Therefore the number of salt moles must be greater than the number of acid moles. In this case most of the glycine is converted to its sodium salt. – Babu May 30 '17 at 1:43
• @bob I also don't think this answer can be correct. The only substance that buffers in the scenario is glycine and the converted sodium salt. Therefore the initial amount of substance of glycine is also the total amount of substance of the buffering components. – Martin - マーチン May 30 '17 at 10:31

It is much more simple. Just make 0.1 M solution of glycine by dissolving 0.75g glycine per 100ml volume (0.1 * 0.1 * 75). In the beginning dissolve it only in about 70 ml water. Adjust the pH of this glycine solution to 10.6 with 1N NaoH. Adjust the volume to 100ml. By my HH calculation it will consume about 900 micro liters of 1N NaoH. You can use more dilute NaOH to have a better control while adjusting the pH. The molarity of the final solution will be 0.1N not any fraction.

You can use this approach for making virtually any buffer. The approach is that you will start with the free acid weighed to the desired molarity and adjust the pH upwards.

Starting with a sodium salt of the free acid and adjusting the pH downwards with HCl is not desirable because that would introduce chloride ion into the system in the form of sodium chloride which can interfere with electropheresis.