I'm a dentist, working on a medical device that will make shots at the dentist office hurt a lot less. The device mixes a drug (2% Lidocaine) that has been acidified with $\ce{HCl}$ to pH 3.5 for storage, with $\ce{NaHCO3}$ to bring it to physiological pH (about 7.2) just prior to injection.

The current method is by manually mixing an 8.4% $\ce{NaHCO3}$ solution with the drug in a 1:9 ratio V/V. In medicine, an 8.4% solution typically means 84mg of drug / ml of solution. So, the current protocol of mixing 100 ml of bicarb solution with 900 ml of drug solution, means the doctors are delivering 8400mg of $\ce{NaHCO3}$ /L , which is a LOT more than the 27mg that my current calculations (please see below, and please let me know if I'm off on the wrong track) say they need to do the job. You can see why I was (and am) skeptical of my calculations.

Assuming the calculations below are correct, then the next problem is:

If I want to deliver the $\ce{NaHCO3}$ in the form of a fluid (whether a solution, slurry, or any other way to make it flow), what is the best way to do that, and how much $\ce{NaHCO3}$ can I fit into a vessel with volume of 35 cubic millimeters (pretty small, and all the room I have available in my device)? In other words, what is the maximum concentration of $\ce{NaHCO3}$ that will stay in solution at a standard temperature and pressure with no head space in the vessel? Or, can I make a slurry that will hold an even higher concentration that remains fluid, and what would that concentration be? I only need to fit 27mg/L x 2ml = 0.054mg of $\ce{NaHCO3}$ into the space, and it would be a lot easier to manufacture if it were a liquid rather than a powder.

Also, I'm in need of some physical testing of the device to confirm that mixing has properly occurred at the tip of the needle when the prototype is built in a few weeks. There is some thought that we could use excess $\ce{NaHCO3}$ in a chamber, and have bypass mixing occur as the drug passes through the chamber and into the needle, instead of trying to get the whole volume (2ml) mixed prior to moving any drug out of the needle. If there is someone on here that might be interested in performing those validation experiments to help pass FDA gates, then please let me know how to find you. Thanks, Scott

The chemical reaction: $\ce{HCl + NaHCO3 → NaCl (aq) + CO2(g) + H2O(l)}$

The initial hydrogen ion concentration is: (pH = 3.5) [H+]i = 10-3.5 = $\pu{3.16228 x 10^-4 M}$

The required final hydrogen ion concentration is: (pH = 7.2) [H+]f = 10-7.2 = $\pu{6.309573 x 10^-8 M}$

moles $\ce{H+}$ reacted = moles of $\ce{HCl}$ reacted = $(\pu{3.16228 x 10^−4} − \pu{6.309573 x 10^−8}) = \pu{3.1616467 x 10^−4 mol}~\ce{HCl}$

moles of $\ce{NaHCO3}$ required = moles $\ce{HCl}$ reacted = $\pu{3.1616467 x 10^−4 moles}$

mass of $\ce{NaHCO3}$ required = $\pu{3.1616467 x 10^−4 moles}~\ce{NaHCO3} \times \pu{84g}~\ce{NaHCO3} / \pu{1 mole}~\ce{NaHCO3} = \pu{0.027 grams}$

  • 1
    $\begingroup$ I'd also wonder about the solubility of Lidocaine as a function of pH and ionic strength. $\endgroup$
    – MaxW
    Commented Dec 20, 2017 at 3:09
  • $\begingroup$ Given that Lidocaine is protonated at pH 3.5, don't you need more base to neutralize the Lidocaine than to neutralize the HCl? // 2% solution is 20 mg/ml or 20 grams per liter. MW of Lidocaine =234, moles = 20/234 = 0.085 moles/liter. 1/6 of that is 0.0142 moles per liter. $\endgroup$
    – MaxW
    Commented Dec 20, 2017 at 3:17

2 Answers 2


The approach I would suggest is a bit different.

First I would calculate the nominal concentration of HCl (i.e. the number of moles of HCl added, divided by the total volume of the solution after the addition, regardless of what happens to the resulting chloride ions and protons) in your initial solution, using all the necessary equations (with the approximation that activity equals concentation, although the figures are a bit large here).

1) dissociation constant of lidocaine:

$K_{a,lido}=\frac{[L] \cdot [H^+]}{[LH^+]}$

2) the total nominal concentration of lidocaine must be the sum of all its forms:

$C_{lido}= [L]+[LH^+]$

3) charge balance:

$[Cl^-] + [OH^-] = [H^+]+[LH^+]$

4) the total concentration of chloride is the same as the nominal concentration of HCl, as the dissociation is complete and $Cl^-$ does not react with anything else:

$C_{HCl}= [Cl^-]$

5) the ion product of water:

$K_w=[H^+] \cdot [OH^-]$

The system of these 5 equations can be solved for $C_{HCl}$, after eliminating all chemical species (except $[H^+]$, so in total 4 unknowns):

$C_{HCl}= [H^+] - \frac {K_w}{[H^+]} + \frac {C_{lido}}{1+ \frac {K_{lido}}{[H^+]}}$

Plugging in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-3.5} M, K_w = 10^{-14} M^2$

results in:

$C_{HCl}= 0.08566 M$

i.e. to reach a pH of 3.5 starting from freebase, we used just slighly more than the stoichiometric amount of $HCl$. Here we are assuming as well that the added volume of HCl has not changed the nominal concentration of lidocaine very much. If it has, then you need to account for that.

Now, when you add sodium bicarbonate, all the equations above except 3 are still applicable; you need to add the dissociation constants for carbonic acid (I'll consider it like hydrated $CO_2$, and that none of it is escaping as gas, to keep things simple) and hydrogenocarbonate, and the mass balances for sodium and the various carbonate species.
Here they are:

$K_{a,carbonic acid}=\frac{[HCO_3^-] \cdot [H^+]}{[H_2CO_3]}$

$K_{a,hydrogcarbonate}=\frac{[CO_3^{2-}] \cdot [H^+]}{[HCO_3^-]}$

$C_{bicarb}= [Na^+]$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}]$

The charge balance now includes more ions:

$[Cl^-] + [OH^-] + [HCO_3^-] + 2 \cdot [CO_3^{2-}] = [Na^+]+[H^+]+[LH^+]$

We have now 9 equations (5-1+5). Like before, eliminating all chemical species except $[H^+]$ (i.e. 8 unknowns), gives a single expression. It is a bit complicated, but essentially it contains only $[H^+]$, $C_{bicarb}$ and $C_{HCl}$, and known constants.

In our case, as we know what the nominal concentration of $HCl$ is from the previous calculation, and the final pH we want to reach (7.2), we just need to solve for $C_{bicarb}$ and plug in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-7.2} M, C_{HCl}=0.08566 M, K_{a,carbonic acid}=4.3 \cdot 10^{-7}, K_{a,hydrogcarbonate}=5.6 \cdot 10^{-11} $

resulting in:

$C_{bicarb}=0.1142 M$

So for each litre of your initial solution you should add 0.1142 mol of $NaHCO_3$, which is 9.59 g.

This is closer to the 'official' quantity you mentioned (even quite a bit more in fact), although I have trouble following all the figures. Maybe they did this experimentally and found that some $CO_2$ escapes, making the pH rise without the need for more base. Just a hypothesis.

In a way, even without all these involved calculations, one could say: if you have a 0.085 M solution of lidocaine hydrochloride and want to react all the $HCl$ with sodium bicarbonate, use $0.085 mol/L \cdot 84 g/mol = 7.14 g/L$ of $NaHCO_3$. The excess used in your official method compared to this figure is probably due to the slight excess of $HCl$ in the initial solution, and the need to take the pH closer to the desired 'physiological' value you mentioned.

EDIT: addendum to answer further questions by the OP

The OP asked what would happen by adding more sodium bicarbonate.

The answer is: as long as there is no solid precipitate of sodium bicarbonate (and always assuming that lidocaine stays in solution), the above equations remain valid; as soon as the nominal concentration of sodium bicarbonate is sufficiently high for it to precipitate, new equations apply, where all the above equations still apply, except for a correction of the mass balances related to $C_{bicarb}$, in particular by adding the contribution of the solid:

$C_{bicarb}= [Na^+] + [NaHCO_3]_s$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}] + [NaHCO_3]_s$

(where $[NaHCO_3]_s$ represents the solid salt expressed in the units of concentration, for homogeneity with the other terms).

And the equation for the solubility product of $NaHCO_3$ must be added:

$K_{s,bicarb}=[Na^+] \cdot [HCO_3^-]$

From the data table in this site it seems that $K_{s,bicarb}=10^{-0.55} M^2 = 0.2818 M^2$ (although from my own calculations based on a solubility of 96 g/L at 20°C it should be $1.277 M^2$, but OK, I will use the 'literature' value).

First, the limiting value of $C_{bicarb}$ after which residual solid bicarbonate is observed at equilibrium must be calculated.

This can be done by solving (numerically) the system of all 10 equations described above for all the chemical species plus $C_{bicarb}$, after setting $[NaHCO_3]_s=0$. The result is:

$[C_{bicarb}=0.5467 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

meaning that this system is only valid for $C_{bicarb} > 0.5467 M$.

Then the system can be solved using any value of $C_{bicarb}$ greater than the above threshold, and, interestingly, it gives always the same values, except for $[NaHCO_3]_s$, that increases accounting for the undissolved salt.

$C_{bicarb}=0.7 M$

$[[NaHCO_3]_s=0.1533 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

$C_{bicarb}=7 M$

$[[NaHCO_3]_s=6.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

$C_{bicarb}=70 M$

$[[NaHCO_3]_s=69.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

So it appears that, once solid sodium bicarbonate is present, the system is determined in a specific state, where all chemical species have a fixed concentration, regardless of how much solid there is.

In particular, the pH seems to be fixed at $7.60$. This is significantly lower than the pH of a saturated solution of $NaHCO_3$ ($8.31$) in water, probably because of the $H_2CO_3$ liberated in the reaction with $HCl$ (which, we are assuming, is remaining in solution).

  • $\begingroup$ If the NaHCO3 were to be delivered in excess to the acidified drug solution, am I correct in saying that the pH would wind up in the 9.6 range? And I'm guessing that wouldn't feel too good for the patient either? $\endgroup$
    – JSK28031
    Commented Dec 20, 2017 at 14:28
  • $\begingroup$ And would that free up even more unionized Lidocaine for pharmacological action on the nerve membrane? $\endgroup$
    – JSK28031
    Commented Dec 20, 2017 at 14:36
  • $\begingroup$ As for the second question: sure, the higher the pH, the more non-protonated lidocaine there will be in solution (the dissociation constant of lidocaine is always holding true), although after a certain pH the difference becomes practically insignificant, I reckon. As for what will happen in the patient's gums, no idea, it's more a physiology question. I'd have thought that the body is pretty good at buffering whatever you inject in it, but how fast that will happen in different tissues, and the effect of injecting a rather alkaline liquid parenterally are outside my field of expertise. $\endgroup$ Commented Dec 20, 2017 at 17:25
  • $\begingroup$ For the first question, I removed my earlier comment, which was imprecise, and I will edit the original answer instead. $\endgroup$ Commented Dec 21, 2017 at 15:48
  • $\begingroup$ Okay, this next question is more chemical engineering than chemistry, I suppose, but I think (hope, actually) your answer suggests: If we take 2ml of 2% lidocaine HCl solution (contains 40mg of lidocaine and 0.08566M of HCl to reach pH 3.5), and then push it at a rate of about .25ml/second through a "mixing" chamber of volume 0.05 ml that contains excess NaHCO3 (let's say, 38mg, which is about 2x what's necessary), the output from the mixing chamber should remain < pH 7.69, and the reaction should proceed rapidly enough to get the effluent solution > pH 7.2? $\endgroup$
    – JSK28031
    Commented Dec 22, 2017 at 2:47

One thing to consider: the Lidocaine is itself a base that gets protonated in the acidified solution. You may need to convert some of the Lidocaine back to its deprotonated, basic form, along with neutralizing the "free" acid that the pH meter sees, when you add the bicarbonate. According to the comments the pKa of Lidicaine is about 7.9 (comments) or 7.7 (https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.frca.co.uk/documents/anaes.5.5.158.pdf&ved=0ahUKEwiEs6nPvZfYAhUl7IMKHc7FBe0QFggmMAE&usg=AOvVaw0RhVYVPEby1Mgu8nIUy48v), so to get to pH 7.2 you must in fact deprotonate at least one sixth of the Lidocaine.

  • 1
    $\begingroup$ The literature says the pKa of Lidocaine is 7.9, and as you point out, at pH 3.5 it exists predominately (99.996%) in it's protonated form, which has no ability to cross the lipid membrane of nerve channels.The body can do this over time, but the NaHCO3 makes it almost instantaneous. This is another good reason to buffer anesthetic, to save time getting numb. $\endgroup$
    – JSK28031
    Commented Dec 20, 2017 at 1:15
  • $\begingroup$ So we deprotonated a limited amount of the Lidocaine, about 1/6 of it. $\endgroup$ Commented Dec 20, 2017 at 1:51
  • 1
    $\begingroup$ Yes, that is consistent with what I have read. Most of it remains pharmacologically unavailable, even after the body fluids bring the ambient pH to about 7.4 $\endgroup$
    – JSK28031
    Commented Dec 20, 2017 at 2:05
  • $\begingroup$ Still need to add it into the calculations though. $\endgroup$ Commented Dec 20, 2017 at 2:17
  • 1
    $\begingroup$ No argument there, but if I thought I could do the calculations (even the basic, no pun intended, calculation), I wouldn't have posted the question in the first place. Do they look correct to you? Or did I mess things up somewhere? I can't believe that the convention in medicine is so far away from what I calculated, it makes me think I missed something big, on order of 100 times or more. $\endgroup$
    – JSK28031
    Commented Dec 20, 2017 at 2:22

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